Now, what about updates? We'll use ugh (and suppress the π¬h that should be there) as shorthand for the function that maps (m,b) over Θ(π¬h∙πpa) to (c(m|h),b+m(0★hg)) in Ma(F(πpa)) (or the nirvana-free or sur variant of this), and also use ugh as a function from belief functions to belief functions (just map all the sets through)

Lemma 27:When updating, the closure adds no nirvana-free points that weren't present originally if Nonemptiness, Nirvana-free upper-completion and closure holds originally (works in the sur-case too)

Proof sketch: We take a sequence Mn limiting to M, and then take a preimage point of Mn, go to a minimal below it, find a limit point in our original set by Compactness, and map it back through the update, getting a point below M. Then, we find what we need to add to that to get M, and find something above our limit point that maps to M, so we didn't actually need closure anyways because we made M as an image of a nirvana-free point present in the original set.

Proof: Fix a sequence Mn in ugh(Θ)(πpa) (but without the defining closure part in the end) that limit to M which is nirvana-free.

Every Mn has a preimage point M′n∈Θ(π¬h∙πpa) with no nirvana off-h. For each M′n, find a minimal point M′lon below it, which have a λ⊙+b⊙ bound by bounded-minimals, so we can find a convergent subsequence limiting to M′lo (actually, might not be minimal, still a limit of minimal points, though). Shoving the M′lon (and the limit point) back through the update (which is a continuous function), we get a sequence Mlon limiting to Mlo (the thing you get from pushing M′lo through the update).

Since M′n lies above M′lon (upper-completion ordering), then updating preserves that property, because the update function is linear. Thus, all the Mlon lie below their corresponding Mn. Now, we can invoke Lemma 16 to conclude that Mlo lies below M. It lies below a nirvana-free point, so M′lo is nirvana-free as well. Now, we just need to show nirvana-free upper-completion because M=Mlo+M∗.

We can take M′lo and add on (m∗,b∗) (extend the measure back to the original domain by sticking an h prefix on everything, and saying that the measure is 0 everywhere else), making an a-measure that's in Θ(π¬h∙πpa), by nirvana-free uppper-completion there. By linearity, and the update not affecting (m∗,b∗) (it's 0 outside of h so the g outside of h doesn't get to contribute anything to the b term when we update), updating M′lo+(m∗,b∗) makes Mlo+M∗=M. So, if a nirvana-free point appears post-update (with closure), then it'll appear post-update (without closure).

First, (pr(π¬h∙πhipa),(π¬h∙πlopa)∗(m))(0★hg)=m(0★hg) This is because the projection down doesn't change the measure at all outside of h, and we're evaluating a function that's 0 inside h. So, that takes care of the b term. Also, projection preserves the b term, so our desired equality is:

For the measure term, the first is "restrict to on-h histories, clip off the h prefixes, and project down", and the second is "project down the on-h histories accordingly, then restrict to on-h histories and clip off the h prefix", which are obviously equal.

Proposition 9:For causal, surcausal, pseudocausal and acausal hypotheses, updating them produces a causal, surcausal, pseudocausal or acausal hypothesis as long as renormalization doesn't fail.

Proof sketch: What we can do is consider the "raw update", show that it preserves all nice properties except for renormalization, and then show that the renormalization terms in the update are the proper renormalization terms to use. Thus, we'll define our raw update ugh(Θ) via: ugh(Θ)(πpa) is: Θ(π¬h∙πpa)∩NF off h, mapped through the following function: (m,b)↦(c(m|h),b+m(0★hg)) And then you take the closure of the resulting set at the end.

So, we take our partial policy πpa and glue it to the off-h-partial policy π¬h, go to that part of the original belief function, strip off everything that has nirvana off-h (for Murphy shall not select those, and properly, that should make the b term infinite post-update), slice off the part of the measure off-h, strip off the h prefix from those histories, and go "ok, our utility function is g, let's take the expected utility off-h and fold it into the b term"

If we can get all nice properties but renormalization, we're good, just appeal to Lemma 24. As for showing the conditions: We're in for something almost as bad as one of the directions of the Isomorphism theorem. Nonemptiness, closure, and convexity are trivial, upper completion, pseudocausality, and bounded-minimals are easy and the extreme point condition and causality are moderately tricky.

For the extreme point condition, Step 1 is establishing that ufh(Θ)(πpa)∩NF equals the closed convex hull of nirvana-free projections from above by an argument that makes sense when you sketch it out but may be difficult if you don't have it sketched out, step 2 is using Hausdorff-continuity and Lemma 20 to turn it into an ordinary convex hull, and finally, arguing that a nirvana-free exteme point must have come from a nirvana-free point from above via step 2.

For causality, we can (for the most part) just go back to Θ, get an outcome function there, and map it back through the update to get an outcome function, the hard part is netting the limit points, which requires a limit of outcome functions. But because we want a countable product of sets to get sequential compactness from Tychonoff, we have to work with stubs, which adds some extra complexity.

Hausdorff-continuity is just hellishly hard, we need to show that the preimages of the sets post-update are the updates of the preimages of sets pre-update, and then combine that with some fancy work with minimal points and upper completions and using two different characterizations of uniform continuity at once via Lemma 15, and a touch of functional analysis. There's way too many interacting points and sets in this one.

But easily the biggest grind is Consistency. We have 4 subset directions to show, each of which requires their own separate fancy argument, and two of them require splitting into a nirvana-containing/causal case and a nirvana-free case, so it's a 6-part proof. A good chunk of complexity arises because we have to take closure in the nirvana-containing case, an issue which goes away if we just let Nirvana be 1 reward forever. Let's begin.

Condition 1: Nirvana-free Nonemptiness:

This is trivial. Just pick a nirvana-free point in Θ(π¬h∙πpa) by nirvana-free nonemptiness, and update, to get one in ugh(Θ)(πpa).

Conditions 2,3: Closure, Convexity:

Closure is a tautology since we took the closure. For convexity, the closure of a convex set is convex, and ugh is a linear function, so it maps convex sets to convex sets.

Condition 4: Nirvana-free Upper-Completeness:

First, invoke Lemma 27 to see that all nirvana-free points must have been present in the raw ugh(Θ(π¬h∙πpa)) set to begin with, without the closure. What we want is that, if M′=M+M∗, and M lies in the raw updated set and is nirvana-free, and M′ is a nirvana-free a-measure, then M∗ lies in the updated set as well.

Find a M′′ that maps to M after updating. It must be nirvana-free, because the nirvana either occurs without h as a prefix (which is forbidden because all that stuff gets clipped off and doesn't get pushed through the update), or the nirvana occurs with h as a prefix, but then it'd show up in the measure component of M post-update, contradicting its nirvana-freeness. Now, we can consider M′′+(m∗,b∗) (basically, M∗, but we take the measure back by sticking an h prefix on everything, and saying that it's 0 off-h). This is present in Θ(π¬h∙πpa), by nirvana-free upper completion. By linearity of updating, and m∗ having no measure in any off-h area where it'd get picked up by g, this updates to M+M∗, witnessing that M′ lies in the image of the update, so we get nirvana-free upper completion.

Condition 5: Bounded-Minimals:

For bounded-minimals, we can pull the Lemma 16 trick of taking our M of interest that Mn limit to, taking a preimage M′n for each Mn, finding a minimal M′lon below each M′n (which obeys a λ⊙+b⊙ bound and also has no nirvana off-h), getting a limit point M′lo (still no nirvana off-h) by compactness, and pushing the sequence through the update, to get a sequence Mlon below Mn limiting to Mlo which is below M (Lemma 16) Now, we just have to check up on the λ+b values of our Mlon sequence, and show that they respect the λ⊙+b⊙ bound, to transfer this to Mlo. The raw update deletes measure from off-h, and assigns it the value that g does, which is 1 or less, so any increase in b correspond to an equal-or-greater decrease in λ, so the Mlon all obey the λ⊙+b⊙ bound as well. Thus, the limit point Mlo obeys the bound, and it's below our original M, so any minimal must obey the λ⊙+b⊙ bound.

Condition 7: Consistency:

This is going to be extremely tedious and difficult to show, it's a 6-part proof. The first 3 parts are devoted to showing that ugh(Θ)(πpa)=¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π)))

Part 1 is showing ugh(Θ)(πpa)⊆¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π))) In the nirvana-free pseudocausal/acausal case.

Let M be in ugh(Θ)(πpa). By Lemma 27, since we're working in the nirvana-free case, we didn't need to take the closure, it won't add any points that aren't there anyways. So, M has a preimage point M′∈Θ(π¬h∙πpa) that maps to it. By consistency for Θ, M′ lies in the closed convex hull of projections of policies down from above, so there are points in the convex hull of projections of policies that are arbitrarily close to M′, fix some sequence M′n of points in the convex hull of projections down from policies above that limit to M′. Mapping these through the raw update (which is continuous) we get a sequence Mn of points in ugh(Θ)(πpa) that limit to M.

All these policies above (π¬h∙πpa) have the form (π¬h∙π). So, M′n can be written as a mix of finitely many M′i,n, which are the projections of M′∞i,n from above, in policies. Update those, getting points M∞i,n in ugh(Θ)(π). These project down to Mi,n, which mix back together to make... Mn. This is because of Lemma 28, that update-then-project equals project-then-update. Also, mix-then-project equals project-then-mix. Remember, Mn is made by: "Project M′∞i,n down to make M′i,n, then mix to make M′n, then update."

So, we can go project-mix-update equals mix-project-update equals mix-update-project equals update-mix-project equals update-project-mix, which is the process "update the M′∞i,n to M∞i,n, project down to Mi,n, mix to Mn"

The first equality is linearity of projection, the second equality is Lemma 28, the third equality is linearity of updating, the final equality is linearity of projection again.

Anyways, taking stock of what we've done, we have a sequence Mn limiting to our M of interest, and every Mn is crafted by taking points from finitely many ugh(Θ)(π), projecting them down, and mixing them. Therefore, our M∈ugh(Θ)(πpa) lies in the closed convex hull of projections down from above.

Part 2: we'll show this again, but in the nirvana-containing case, where we'll leverage causality.

Fix a M∈ugh(Θ)(πpa) (with closure). There's a sequence Mn that limits to it, that lies in the same set, but without closure, so we can take preimage points M′n∈Θ(π¬h∙πpa) that update to make Mn. By causality, fix some arbitrary policy above (π¬h∙πpa), which can be expressed as (π¬h∙π), where π≥πpa. Anyways, we can take M′n, and use causality to get an outcome function of, to get a M′∞n∈Θ(π¬h∙π) that projects down to M′n. We don't have to worry about nirvana off-h, because M′n already specifies everything that happens off-h and it says no nirvana occurs in that case. So, M′∞n can be updated to make a M∞n in ugh(Θ)(π). By Lemma 28, this must project down to Mn. So, all our Mn lie in the projection of ugh(Θ)(π), and since M is a limit point of that sequence, it must lie in the closed convex hull of projections.

And we've shown that ugh(Θ)(πpa)⊆¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π)))

And have taken care of 2 of our 6 parts. Now for the reverse direction, that

Thankfully, this can be done with a general argument that isn't sensitive to the presence of Nirvana.

Part 3: Fix a M in the closed convex hull, which has a sequence Mn limiting to it that's in the convex hull of projections down from above. the Mn shatter into finitely many Mi,n, which are projections of M∞i,n∈ugh(Θ)(πi). Now, these aren't necessarily preimage points, they may have been added in the closure. Thus, we can perturb by 2−n or less if needed to make a M′∞i,n which does have a preimage point. Projecting these down to M′i,n and mixing, crafts a M′n point that is within 2−n of Mn (remember, projection doesn't expand distance), so the sequence M′n still has M as a limit point (it gets arbitrarily close to a sequence that gets arbitrarily close to M). If we can show that all the M′n lie in ugh(Θ)(πpa), then by closure, we'll get that M lies in the same set so we're done.

Ok, so we have M′∞i,n∈ugh(Θ)(πi), that project down and mix to make M′n, and importantly, we crafted them so they're produceable without closure. Thus, they have preimage points M′′∞i,n∈Θ(π¬h∙πi) (that lack nirvana off-h) Project them down to make M′′i,n∈Θ(π¬h∙πpa), and mix them to make a M′′n in the same set (which still lacks nirvana off-h), and this updates to make M′n via Lemma 28, as we'll show shortly.

Starting with the M′′∞i,n, we know that update, project, mix equals M′n via going M′∞i,n, M′i,n, M′n. Then, update-project-mix equals project-update-mix equals project-mix-update, which is the path we took. Therefore, all the M′n lie in ugh(Θ)(πpa), which is closed, so M (arbitrary in the closed convex hull of projections) lies in the same set, establishing the reverse subset direction and thus equality,

Part 4: Now that we're halfway done,let's look at the "intersection of preimages of stubs from below" direction of consistency, ugh(Θ)(πpa)⊆⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa)). If we ignore the closure part and work with the raw update set sans closure, we can fix a M in ugh(Θ)(πhipa), take a preimage point in Θ(π¬h∙πhipa), project it down to Θ(π¬h∙πlopa) by consistency, then update it to get exactly the projection of M (again, Lemma 28) Then, when we take the closure, we can just take our M in the closure, fix a sequence in the raw update set sans closure Mn that limits to M, project down, getting M′n in the raw update set ugh(Θ)(πlopa) sans closure, and then the limit point M′ lies in ugh(Θ)(πlopa) by closure, and by continuity of projection, M′ is the projection of M.

Since the sets get bigger as you go down, we can invoke Lemma 6 to swap out the intersection of preimages of all stubs below you, for the intersection of preimages of stubs of the form πnpa, this will be important later.

Now, it's trivial to show that ugh(Θ)(πpa)⊆⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa)) because we've established that projecting down makes a subset, and projection commutes, so any M∈ugh(Θ)(πpa) projects down into ugh(Θ)(πnpa) for all n.

All that's left now is the reverse subset direction,

ugh(Θ)(πpa)⊇⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa))

Sadly, this one will require us splitting into the nirvana-containing (and thus causal) cases and the nirvana-free cases, and it's a really difficult one to show.

Part 5: Let's address the nirvana-free case, we'll use a nifty trick to control the size of the preimage points we select.

Ok, let's say you have a M with some λ1 and b1 value. And you take M′ that's a preimage point, but its λ and b values are just... waaay too high. We want to have a preimage point with reasonable values, in order to apply bounding arguments. What you do, is find a minimal-point Mmin below M′, so M′=Mmin+M∗. Now, what you do, is swap out M∗ ie (m∗,b∗), for (m∗|h,b∗+m∗(0★hg)). This is an sa-measure, because

b∗+m∗(0★hg)+(m∗|h)−(1)=b∗+m∗(0★hg)+m∗−(1★h0)

≥m∗−(0★hg)+m∗−(1★h0)=b∗+m∗−(1★hg)≥b∗+m∗−(1)≥0

Now, consider updating Mmin+(m∗|h,b∗+m∗(0★hg)) instead (it's an a-measure, it has less negative parts than M∗, and is present by nirvana-free upper-completion). This gets you the update of Mmin, plus... (c(m∗|h),b∗+m∗(0★hg)) (remember, 0 measure off-h). Which is the exact same thing you'd get by updating M∗, so when we updated our new sum, we hit M exactly.

However, this sum is special, because we can stick some decent bounds on its λ and b value! For starters, its b value is less than b1 (updating only adds on b-mass, and it updates to M). And as for the λ value... well, Mmin has its λ bounded above by λ⊙ (of the original Θ) due to being a minimal point. And in the worst-case, all of the measure in M came from the thing we added, so m∗|h has a measure of λ1 or less. So our bound on the λ value is λ⊙+λ1.

Armed with this knowledge, we can begin to prove the last bit of consistency in the nirvana-free case. Take a M in the intersection of preimages. It projects down to make Mn in ugh(Θ)(πnpa). Projection preserves λ and b, so they all have the same λ1,b1 bounds. Because we don't need to close in the nirvana-free case, we get a preimage point of M′n in Θ(π¬h∙πnpa) From our earlier considerations, we can always pick M′n such that its λ is ≤λ⊙+λ1, and its b is ≤b1, although we'll be using a bound of max(b1,b⊙).

Now, we're going to have to be extremely careful here. Let the point M′n,j be defined as:

If j<n, then M′n,j is some arbitrary point in Θ(π¬h∙πnpa), with λ equal to or below λ⊙+λ1, and b equal to or below max(b1,b⊙), which always exists by all minimal points obeying the λ⊙+b⊙ bound.

If j=n, then M′n,j=M′n.

If j>n, then M′n,j=pr(π¬h,πjpa),(π¬h,πnpa)∗(M′j)

Then, the tuple of M′n,j for all n is a point in:

∏n(Θ(π¬h∙πnpa)∩{(λμ,b)|λ≤λ⊙+λ1,b≤max(b1,b⊙)})

Equipped with the product topology. In particular, this is a product of compact sets, so by Tychonoff's theorem, it's compact. Thus, we can get a convergent subsequence of the tuples. On this subsequence, all the M′n,j converge to a limit point M′n,∞, regardless of n.

Also, M′n,∞ projects down to M′m,∞ if n≥m, because for large enough j, the projection of M′n,j will always be M′m,j, and by continuity of projection, the projection of M′n,∞ must be M′m,∞

Ok, so we've got an infinite sequence of M′n,∞ for all n that all project down onto each other. Another nice feature is that updating M′n,∞ produces Mn. This is because, when j climbs high enough, M′j,j projects down to M′n,j, and M′j,j is just M′j which updates to Mj, which projects down to Mn. By Lemma 28, update-then-project equals project-then-update, so M′n,j must update to Mn, for all sufficiently high j. The preimage of a single point is closed, so past a certain point, the M′n,j are wandering around in the preimage of Mn, so M′n,∞ also updates to Mn.

Now, our next step is, does the M′n,∞ sequence in Θ(π¬h∙πnpa) pick out a single point M′ in Θ(π¬h∙πpa) that projects down accordingly? Yes it does. Just intersect all the preimages of single points, they're nested in each other and compact so the finite intersection property holds, and if the intersection wasn't composed of a single point, you'd have two distinct points with a difference at some finite time, but projecting down to any finite time the two distinct points are identical, so there can only be a single point in the intersection. Further, it must lie in Θ(π¬h∙πpa), because you can project it down to M′n,∞ in Θ(π¬h∙πnpa) for any n, which, by consistency for Θ, you can also project down to Θ((π¬h∙πpa)n) (projecting down further), so it's present in the intersection of all the preimages, certifying that it's in the appropriate set.

Now, finally... does M′, when updated, produce M, certifying that the point in the intersection of preimages is also in the raw update set? Well, let's say it didn't. Then we get a M′′ that's not equal to M, so projecting down to some finite n should suffice to observe that. However, projecting M′′ and M down produces... Mn. This is because of Lemma 28, update-then-project equals project-then-update. Projecting M′ down makes M′n,∞, which updates to Mn.

So, no finite stage suffices to observe the difference between the updated form of M′ and M itself, so they must be identical, certifying ugh(Θ)(πpa)⊇⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa)) for the nirvana-free case.

Part 6: Let's move to the nirvana-case where we can leverage causality. We'll be showing this in a rather nonstandard way. We're going to pick a π≥πpa, and show that our M of interest in the intersection of preimages can be written as a limit of points projected down from ugh(Θ)(π), establishing that M lies in the closed convex hull of points from above, which we've already shown equals ugh(Θ)(πpa).

Ok, so M is in the intersection of preimages. Project it down to all the ugh(Θ)(πnpa), getting a batch of points Mn from them. This is the raw update set, so within 2−n or less distance from Mn, there's a M′n in the raw update sans closure, which has a preimage point M′′n that lies in Θ(π¬h∙πnpa).

Now, pick some arbitrary policy above π¬h∙πpa, which can be written as π¬h∙π. Moving on even further, by causality, we can get a point M′′∞n∈Θ(π¬h∙π) that projects down to M′′n. Update M′′∞n to get a M′∞n∈ugh(Θ)(π), which then (by our earlier thing about how a set equaled the closed convex hull of projections down from above), projects down to a M′hin∈ugh(Θ)(πpa).

Now, we can ask whether the sequence M′hin limits to M itself. ugh(Θ)(πpa) is closed, so this would certify that M lies in the appropriate set.

First, observe that the projection of M′hin down to πnpa is M′n. This is by Lemma 28, update-then-project equals project-then-update. M′′∞n projects down to M′′n, which updates to M′n, so M′n must be what you get by updating M′′∞n to M′∞n, and projecting down to πpa (making M′hin), and projecting that down to πnpa.

Now, because projection preserves the b term, and M′hin projects down to M′n which is within 2−n of Mn (not much of a difference in the b terms), and Mn has the same b term as M, we can certify convergence of the b term at least. Now for convergence of the measure components. Again, M′hin projects down to M′n which is within 2−n of Mn (not much difference before timestep n, shrinking increasingly low), and Mn perfectly mimics what M does before timestep n. So, M′hin behaves increasingly closely to M for everything before time n, which increases without bound. Increasingly close matches on increasingly large initial segments of what happens mean that M′hin must limit to M itself, certifying that M lies in ugh(Θ)(πpa) for the causal cases.

That's the last bit we needed! We're finally done with consistency now. This just leaves the hausdorff-condition and the extreme-point condition and pseudocausality and causality.

Condition 9: Hausdorff-continuity:

What we need to do for our setup to even approach this is to show that updating the preimage of the nirvana-free part of Θ(π¬h∙πpa), produces exactly the preimage of the nirvana-free part of ugh(Θ)(πpa).

One direction, we can get easily. If you fix a M′∞ in the preimage of the nirvana-free part of Θ(π¬h∙πpa), it projects down to a M′∈Θ(π¬h∙πpa)∩NF, that updates to a M∈ugh(Θ)(πpa), then by Lemma 28, project-then-update equals update-then-project, so M′∞ must update to a M∞ that projects down to M, certifying that updating the preimage of the nirvana-free part of Θ(π¬h∙πpa) produces a subset of the preimage of the nirvana-free part of ugh(Θ)(πpa).

In the other direction, fix a M∞ in the preimage of the nirvana-free part of ugh(Θ)(πpa). It projects down to a M in ugh(Θ)(πpa)∩NF, and by Lemma 27, M wasn't introduced in the closure, so it has a preimage point M′∈Θ(π¬h∙πpa)∩NF.

Now, how do we extend M′ to craft a M′∞ that updates to M∞? Well, we can split into two parts. What happens on-h, and what happens off-h? For the off-h part, the post-update part has everything folded into the b term, while the pre-update part has an actual measure specified everywhere. Thus, our M′∞ should have the same off-h part as M′ to project down accordingly, so updating folds it into the same b term as M∞ has.

Now, for the on-h part, it's a bit more complicated. M∞ specified what happens for all infinite histories with h as a prefix. However, M and M′ only specify part of that data, but fortunately agree on that part. Thus, for M′∞, you can just extend with the conditional probabilities of M∞, to perfectly mimic it on-h. This makes a M′∞ in the preimage that updates to M∞.

Ok, so the appropriate preimages for Hausdorff-continuity (post-update) are made exactly by updating the preimages for Hausdorff-continuity (pre-update). Now, updating is a continuous linear operator. We're mapping from the Banach space M±((A×O)ω)⊕R to the Banach space M±(h(A×O)ω)⊕R.

Well, this isn't quite right, your actions and observations may vary depending on where you are in history, but the general thing of "restrict to signed measures over infinite histories with h as a prefix" still checks out. Updating is still a continuous linear operator between Banach spaces, by Lemma 8 of Section 1.

Also, all continuous linear operators between Banach spaces are bounded, and thus Lipschitz-continuous at 0, and thus Lipschitz-continuous everywhere due to linearity. So, when we push two points that are only ϵ apart through the update, they're now Cϵ apart at most, where C is a finite constant.

We're going to have a lot of points. Unusually enough, we'll be using the standard formulation of Hausdorff-continuity for our original Θ, that for all ϵ, there's a δ where two partial policies πpa and π′pa that are δ or less apart have (pr∞,πpa∗)−1(Θ(πpa)∩NF∩{≤⊙}) (and the analogous set for π′pa) being only ϵ apart in Hausdorff-distance.

Fixing your ϵ, you're gonna want δ to be low enough to force a ϵC difference between the clipped preimages, and δ<ϵC. It's highly advised to sketch out how our points interact and what sets they're in. A superscript of infinity will be used to denote points in the preimages of the ugh(Θ)(πpa) sets (or Θ(π¬h∙πpa)) (ie, at the infinite levels), and a superscript of "u" specifies post-update while its lack is pre-update.

Anyways, here's our points.

Mu,∞ lies in the preimage of ugh(Θ)(πpa)∩NF, and it's our point that we want to find a point nearby. λ will refer to the λ value of this thing.

Projecting Mu,∞ down to ugh(Θ)(πpa)∩NF makes Mu.

We can find a minimal point below Mu, Mu,min in ugh(Θ)(πpa)∩NF. Mu,min+Mu,∗=Mu.

A nirvana-free point wasn't introduced by the closure, and it has a minimal point in its preimage, so there's a Mmin in Θ(πpa) that updates to Mu,min, and respects the λ⊙+b⊙ bound of Θ.

Let Mlo be defined as Mmin+((mu,∗)−,−(mu,∗)−(1)). We're extending the negative-measure part of Mu,∗ back to its original domain by sticking an h prefix on everything, and saying it's 0 everywhere else. this is an a-measure that lies in Θ(π¬h∙πpa)∩NF∩{≤⊙} (because Mmin respects the λ⊙+b⊙ bound, and the thing that we added has a λ+b value of 0)

Let M be defined as Mlo+((mu,∗)+,bu,∗+(mu,∗)−(1)), it also lies in the same set Updating M makes Mu, because, unpacking M, it's Mmin+Mu,∗, which updates to Mu,min+Mu,∗ which adds up to make Mu.

Our goal now is to explicitly construct a M∞ and Mlo,∞ in the preimage of Θ(π¬h∙πpa)∩NF s.t. they project down onto M and Mlo, Mlo,∞ lies below M∞, and M∞ updates to Mu,∞.

A sufficient way to do this is to make Mlo,∞ and M∞ by, after h, extending the measures further with the conditional probabilities of the measure component of Mu,∞. Extending ((mu,∗)+,bu,∗+(mu,∗)−(1)) with the conditional probabilities of Mu,∞ witnesses that Mlo,∞ lies below M∞. They obviously project down onto M and Mlo.

As for M∞ updating to Mu,∞, the b term and the fragment of the measure that doesn't get ignored by projection down matches because M∞ projects to M which updates to Mu which is the projection of Mu,∞. And, for the fragment of the measure that isn't defined in Θ(π¬h∙πpa), but that must be present on the infinite levels, we copied the conditional probabilities of the measure component Mu,∞, so we've got a match there.

Taking a break from setting up all our damn points for a brief recap, we have a Mlo,∞ that lies in the preimage of Θ(π¬h∙πpa)∩NF∩{≤⊙}, and a M∞ that lies above it (in the preimage of Θ(π¬h∙πpa)∩NF), and it updates to hit Mu,∞ (our original point in the preimage of ugh(Θ)(πpa)∩NF). Now, we can proceed.

So... Mlo,∞ lies in the preimage of Θ(π¬h∙πpa)∩NF∩{≤⊙}. By hausdorff-continuity for Θ and the distance between (π¬h∙πpa) and (π¬h∙π′pa) being below δ because the distance between πpa and π′pa is below δ, and using our earlier thing about how a δ distance means a ϵC difference between the clipped preimages, we can find a point (Mlo,∞)′ in the preimage of Θ(π¬h∙π′pa)∩NF∩{≤⊙} that's that close to Mlo,∞. To go up from Mlo,∞ to M∞ requires adding ((mu,∗)+,bu,∗+(mu,∗)−(1)) (with the measure component extended with the conditional probabilities of the measure component of Mu,∞, obviously).

Also, because the λ value of Mu,∞ is the λ value of Mu, which was made by adding Mu,∗ to an a-measure, an upper bound on the λ value of that a-measure we added onto Mlo,∞ is... λ. Corresponding to the extreme case where all the measure of Mu came from Mu,∗.

Now, we can craft a point (M∞)′ which lies in the preimage of Θ(π¬h∙π′pa)∩NF that's only ϵC+δλ away from M∞. Why? Well, we can start with (Mlo,∞)′, which is only ϵC away from Mlo,∞, and take that positive-measure-thingy we added, and reshuffle the measure on it. With earthmover distance, the δ distance between (π¬h∙π′pa) and (π¬h∙πpa) corresponds to a time-threshold where they start to differ at logγ(δ), and you're moving dirt a γlogγ(δ)=δ difference to account for having to land in the right preimage, and you've got λ at most dirt to move. Then, you just add (Mlo,∞)′ and your reshuffled measure, to get your point (M∞)′. Which is the sum of two components that only differ by ϵC and δλ from the components which sum to make M∞.

Ok, so we have a point M∞ in the preimage of Θ(π¬h∙πpa)∩NF, which updates to Mu,∞ that lies in the preimage of ugh(Θ)(πpa). And a point (M∞)′ in the preimage of Θ(π¬h∙π′pa)∩NF which is (taking into account that δ<ϵC) only ϵC(1+λ) distance away from M∞.

And now we can finish up, because the preimage of ugh(Θ)(π′pa)∩NF is the update of the preimage of Θ(π¬h∙π′pa)∩NF. So, we just update (M∞)′ to get a point (Mu,∞)′ in the preimage of ugh(Θ)(π′pa). And further, the distance between M∞ and (M∞)′ is only ϵC(1+λ) at most. M∞ updates to Mu,∞, and (M∞)′ updates to (Mu,∞)′. And we know that ugh has a Lipschitz constant of C (by being a continuous linear operator between Banach spaces), so Mu,∞ only has a distance of ϵ(1+λ) from a point in the preimage of ugh(Θ)(π′pa).

So, we get Hausdorff-continuity (the Lemma 15 variant).

Condition 8: Extreme Point Condition:

We had to defer this because π¬h∙πstisn't a stub, so we can't use the extreme point condition we had, and instead must regenerate it completely from scratch.

Our first step in this is showing ugh(Θ)(πst)∩NF=¯¯¯¯¯¯¯¯c.h(⋃π≥πstprπ,πst(ugh(Θ)(π)∩NF))

One subset direction is easy, the closed convex hull of projections of nirvana-free stuff must all be in ugh(Θ)(πst) by consistency which we've shown, and all must be nirvana-free. Now for the reverse direction. Let M∈ugh(Θ)(πst)∩NF By Lemma 27, this point wasn't added in the closure, so it has a preimage point M′∈Θ(π¬h∙πst)∩NF. Using all our nice conditions for Θ, we can invoke Lemma 21 to get that M′∈¯¯¯¯¯¯¯¯c.h(⋃π≥(π¬h∙πst)prπ,(π¬h∙πst)(Θ(π)∩NF)), so we can fix a sequence M′n limiting to M where each M′n shatters into M′i,n that came from some M′∞i,n that's nirvana-free and lies in the associated set of a full policy above π¬h∙πst.

Updating the M′n produces a sequence Mn which is nirvana-free, in ugh(Θ)(πst), and limits to M by continuity.

Updating the M′∞i,n into M∞i,n which lie in ugh(Θ)(πi)∩NF, projecting down to get Mi,n, and mixing them, produces Mn, by our usual Lemma 28 argument.

This witnesses that all the Mn lie in c.h(⋃π>πstprπ,πst(ugh(Θ)(π)∩NF))

Thus, M lies in the closed convex hull of projections of nirvana-free stuff from above. What do we do with this? Well, now we can invoke Lemma 20, since we have Hausdorff-continuity proved, to conclude that c.h(⋃π≥πstprπ,πst(ugh(Θ)(π)∩NF)) is closed, so we didn't really need the closed convex hull (which we've already shown is the same as ugh(Θ)(πst)∩NF)

And we now know that ugh(Θ)(πst)∩NF=c.h(⋃π≥πstprπ,πst(ugh(Θ)(π)∩NF))

Now, we can take a minimal extreme nirvana-free point Mex in ugh(Θ)(πst). It must be minimal and extreme and nirvana-free in the original set. If it wasn't minimal in the original set, all minimals below it would be nirvana-free too, witnessing its nonminimiality in the restricted set. And if it wasn't extreme in the original set, then the points that mix to make it must all be nirvana-free too, since it's nirvana-free, so we have a witness of non-extremeness in ugh(Θ)(πst)∩NF.

Ok, so it's extreme and nirvana-free. It must also be extreme in the convex hull set, but, since it can't be produced by mixtures, there's a M∞ in someugh(Θ)(π)∩NF that projects down to Mex, establishing the extreme point condition.

That just leaves causality and pseudocausality.

Condition C: Causality

Ok, we pick a πpa and a point in ugh(Θ)(πpa) Can we make an outcome function for everything that includes our point? By our proof of full causality in the first part of the Isomorphism theorem (finite-to-full direction), this can be done as long as all other conditions are met and we can make an outcome function for any point in any ugh(Θ)(πst). So, let's just establish finitary causality. Fix some πst and some M∈ugh(Θ)(πst).

Since M is in the updated set, there's a sequence Mn that limits to M that we don't need closure to get. There's a λ and b bound on this sequence because it converges, call those bounds λ◯ and b◯. Now, we can take a M′n∈Θ(π¬h∙πpa) that updates to Mn. We can use causality for Θ to get an outcome function for M′n.

We don't have to worry about nirvana-off-h, because M′n has no nirvana off-h, and the projection of M′n down to Θ(π¬h

Here are the previous two posts.

Now, what about updates? We'll use ugh (and suppress the π¬h that should be there) as shorthand for the function that maps (m,b) over Θ(π¬h∙πpa) to (c(m|h),b+m(0★hg)) in Ma(F(πpa)) (or the nirvana-free or sur variant of this), and also use ugh as a function from belief functions to belief functions (just map all the sets through)

Lemma 27:When updating, the closure adds no nirvana-free points that weren't present originally if Nonemptiness, Nirvana-free upper-completion and closure holds originally (works in the sur-case too)Proof sketch: We take a sequence Mn limiting to M, and then take a preimage point of Mn, go to a minimal below it, find a limit point in our original set by Compactness, and map it back through the update, getting a point below M. Then, we find what we need to add to that to get M, and find something above our limit point that maps to M, so we didn't actually need closure anyways because we made M as an image of a nirvana-free point present in the original set.

Proof: Fix a sequence Mn in ugh(Θ)(πpa) (but without the defining closure part in the end) that limit to M which is nirvana-free.

Every Mn has a preimage point M′n∈Θ(π¬h∙πpa) with no nirvana off-h. For each M′n, find a minimal point M′lon below it, which have a λ⊙+b⊙ bound by bounded-minimals, so we can find a convergent subsequence limiting to M′lo (actually, might not be minimal, still a limit of minimal points, though). Shoving the M′lon (and the limit point) back through the update (which is a continuous function), we get a sequence Mlon limiting to Mlo (the thing you get from pushing M′lo through the update).

Since M′n lies above M′lon (upper-completion ordering), then updating preserves that property, because the update function is linear. Thus, all the Mlon lie below their corresponding Mn. Now, we can invoke Lemma 16 to conclude that Mlo lies below M. It lies below a nirvana-free point, so M′lo is nirvana-free as well. Now, we just need to show nirvana-free upper-completion because M=Mlo+M∗.

We can take M′lo and add on (m∗,b∗) (extend the measure back to the original domain by sticking an h prefix on everything, and saying that the measure is 0 everywhere else), making an a-measure that's in Θ(π¬h∙πpa), by nirvana-free uppper-completion there. By linearity, and the update not affecting (m∗,b∗) (it's 0 outside of h so the g outside of h doesn't get to contribute anything to the b term when we update), updating M′lo+(m∗,b∗) makes Mlo+M∗=M. So, if a nirvana-free point appears post-update (with closure), then it'll appear post-update (without closure).

Lemma 28:raw-update-then-project equals project-then-raw-update.Take some (m,b). We want to show that:

prπhipa,πlopa∗(c(m|h),b+m(0★hg))

=(c(pr(π¬h∙πhipa),(π¬h∙πlopa)∗(m)|h),b+(pr(π¬h∙πhipa),(π¬h∙πlopa)∗(m))(0★hg))

First, (pr(π¬h∙πhipa),(π¬h∙πlopa)∗(m))(0★hg)=m(0★hg) This is because the projection down doesn't change the measure

at alloutside of h, and we're evaluating a function that's 0 inside h. So, that takes care of the b term. Also, projection preserves the b term, so our desired equality is:(prπhipa,πlopa∗(c(m|h)),b+m(0★hg))=(c(pr(π¬h∙πhipa),(π¬h∙πlopa)∗(m)|h),b+m(0★hg))

For the measure term, the first is "restrict to on-h histories, clip off the h prefixes, and project down", and the second is "project down the on-h histories accordingly, then restrict to on-h histories and clip off the h prefix", which are obviously equal.

Proposition 9:For causal, surcausal, pseudocausal and acausal hypotheses, updating them produces a causal, surcausal, pseudocausal or acausal hypothesis as long as renormalization doesn't fail.Proof sketch: What we can do is consider the "raw update", show that it preserves all nice properties except for renormalization, and then show that the renormalization terms in the update are the proper renormalization terms to use. Thus, we'll define our raw update ugh(Θ) via: ugh(Θ)(πpa) is: Θ(π¬h∙πpa)∩NF off h, mapped through the following function: (m,b)↦(c(m|h),b+m(0★hg)) And then you take the closure of the resulting set at the end.

So, we take our partial policy πpa and glue it to the off-h-partial policy π¬h, go to that part of the original belief function, strip off everything that has nirvana off-h (for Murphy shall not select those, and properly, that should make the b term infinite post-update), slice off the part of the measure off-h, strip off the h prefix from those histories, and go "ok, our utility function is g, let's take the expected utility off-h and fold it into the b term"

If we can get all nice properties but renormalization, we're good, just appeal to Lemma 24. As for showing the conditions: We're in for something almost as bad as one of the directions of the Isomorphism theorem. Nonemptiness, closure, and convexity are trivial, upper completion, pseudocausality, and bounded-minimals are easy and the extreme point condition and causality are moderately tricky.

For the extreme point condition, Step 1 is establishing that ufh(Θ)(πpa)∩NF equals the closed convex hull of nirvana-free projections from above by an argument that makes sense when you sketch it out but may be difficult if you don't have it sketched out, step 2 is using Hausdorff-continuity and Lemma 20 to turn it into an ordinary convex hull, and finally, arguing that a nirvana-free exteme point must have come from a nirvana-free point from above via step 2.

For causality, we can (for the most part) just go back to Θ, get an outcome function there, and map it back through the update to get an outcome function, the hard part is netting the limit points, which requires a limit of outcome functions. But because we want a countable product of sets to get sequential compactness from Tychonoff, we have to work with stubs, which adds some extra complexity.

Hausdorff-continuity is just hellishly hard, we need to show that the preimages of the sets post-update are the updates of the preimages of sets pre-update, and then combine that with some fancy work with minimal points and upper completions and using two different characterizations of uniform continuity at once via Lemma 15, and a touch of functional analysis. There's way too many interacting points and sets in this one.

But easily the biggest grind is Consistency. We have 4 subset directions to show, each of which requires their own separate fancy argument, and two of them require splitting into a nirvana-containing/causal case and a nirvana-free case, so it's a 6-part proof. A good chunk of complexity arises because we have to take closure in the nirvana-containing case, an issue which goes away if we just let Nirvana be 1 reward forever. Let's begin.

Condition 1:Nirvana-free Nonemptiness:This is trivial. Just pick a nirvana-free point in Θ(π¬h∙πpa) by nirvana-free nonemptiness, and update, to get one in ugh(Θ)(πpa).

Conditions 2,3:Closure, Convexity:Closure is a tautology since we took the closure. For convexity, the closure of a convex set is convex, and ugh is a linear function, so it maps convex sets to convex sets.

Condition 4:Nirvana-free Upper-Completeness:First, invoke Lemma 27 to see that all nirvana-free points must have been present in the raw ugh(Θ(π¬h∙πpa)) set to begin with, without the closure. What we want is that, if M′=M+M∗, and M lies in the raw updated set and is nirvana-free, and M′ is a nirvana-free a-measure, then M∗ lies in the updated set as well.

Find a M′′ that maps to M after updating. It must be nirvana-free, because the nirvana either occurs without h as a prefix (which is forbidden because all that stuff gets clipped off and doesn't get pushed through the update), or the nirvana occurs with h as a prefix, but then it'd show up in the measure component of M post-update, contradicting its nirvana-freeness. Now, we can consider M′′+(m∗,b∗) (basically, M∗, but we take the measure back by sticking an h prefix on everything, and saying that it's 0 off-h). This is present in Θ(π¬h∙πpa), by nirvana-free upper completion. By linearity of updating, and m∗ having no measure in any off-h area where it'd get picked up by g, this updates to M+M∗, witnessing that M′ lies in the image of the update, so we get nirvana-free upper completion.

Condition 5: Bounded-Minimals:

For bounded-minimals, we can pull the Lemma 16 trick of taking our M of interest that Mn limit to, taking a preimage M′n for each Mn, finding a minimal M′lon below each M′n (which obeys a λ⊙+b⊙ bound and also has no nirvana off-h), getting a limit point M′lo (still no nirvana off-h) by compactness, and pushing the sequence through the update, to get a sequence Mlon below Mn limiting to Mlo which is below M (Lemma 16) Now, we just have to check up on the λ+b values of our Mlon sequence, and show that they respect the λ⊙+b⊙ bound, to transfer this to Mlo. The raw update deletes measure from off-h, and assigns it the value that g does, which is 1 or less, so any increase in b correspond to an equal-or-greater decrease in λ, so the Mlon all obey the λ⊙+b⊙ bound as well. Thus, the limit point Mlo obeys the bound, and it's below our original M, so any minimal must obey the λ⊙+b⊙ bound.

Condition 7:Consistency:This is going to be extremely tedious and difficult to show, it's a 6-part proof. The first 3 parts are devoted to showing that ugh(Θ)(πpa)=¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π)))

Part 1 is showing ugh(Θ)(πpa)⊆¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π))) In the nirvana-free pseudocausal/acausal case.

Let M be in ugh(Θ)(πpa). By Lemma 27, since we're working in the nirvana-free case, we didn't need to take the closure, it won't add any points that aren't there anyways. So, M has a preimage point M′∈Θ(π¬h∙πpa) that maps to it. By consistency for Θ, M′ lies in the closed convex hull of projections of policies down from above, so there are points in the convex hull of projections of policies that are arbitrarily close to M′, fix some sequence M′n of points in the convex hull of projections down from policies above that limit to M′. Mapping these through the raw update (which is continuous) we get a sequence Mn of points in ugh(Θ)(πpa) that limit to M.

All these policies above (π¬h∙πpa) have the form (π¬h∙π). So, M′n can be written as a mix of finitely many M′i,n, which are the projections of M′∞i,n from above, in policies. Update

those, getting points M∞i,n in ugh(Θ)(π). These project down to Mi,n, which mix back together to make... Mn. This is because of Lemma 28, that update-then-project equals project-then-update. Also, mix-then-project equals project-then-mix. Remember, Mn is made by: "Project M′∞i,n down to make M′i,n, then mix to make M′n, then update."So, we can go project-mix-update equals mix-project-update equals mix-update-project equals update-mix-project equals update-project-mix, which is the process "update the M′∞i,n to M∞i,n, project down to Mi,n, mix to Mn"

The first equality is linearity of projection, the second equality is Lemma 28, the third equality is linearity of updating, the final equality is linearity of projection again.

Anyways, taking stock of what we've done, we have a sequence Mn limiting to our M of interest, and every Mn is crafted by taking points from finitely many ugh(Θ)(π), projecting them down, and mixing them. Therefore, our M∈ugh(Θ)(πpa) lies in the closed convex hull of projections down from above.

Part 2: we'll show this again, but in the nirvana-containing case, where we'll leverage causality.

Fix a M∈ugh(Θ)(πpa) (with closure). There's a sequence Mn that limits to it, that lies in the same set, but without closure, so we can take preimage points M′n∈Θ(π¬h∙πpa) that update to make Mn. By causality, fix some arbitrary policy above (π¬h∙πpa), which can be expressed as (π¬h∙π), where π≥πpa. Anyways, we can take M′n, and use causality to get an outcome function of, to get a M′∞n∈Θ(π¬h∙π) that projects down to M′n. We don't have to worry about nirvana off-h, because M′n already specifies everything that happens off-h and it says no nirvana occurs in that case. So, M′∞n can be updated to make a M∞n in ugh(Θ)(π). By Lemma 28, this must project down to Mn. So, all our Mn lie in the projection of ugh(Θ)(π), and since M is a limit point of that sequence, it must lie in the closed convex hull of projections.

And we've shown that ugh(Θ)(πpa)⊆¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π)))

And have taken care of 2 of our 6 parts. Now for the reverse direction, that

ugh(Θ)(πpa)⊇¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π)))

Thankfully, this can be done with a general argument that isn't sensitive to the presence of Nirvana.

Part 3: Fix a M in the closed convex hull, which has a sequence Mn limiting to it that's in the convex hull of projections down from above. the Mn shatter into finitely many Mi,n, which are projections of M∞i,n∈ugh(Θ)(πi). Now, these aren't

necessarilypreimage points, they may have been added in the closure. Thus, we can perturb by 2−n or less if needed to make a M′∞i,n whichdoeshave a preimage point. Projecting these down to M′i,n and mixing, crafts a M′n point that is within 2−n of Mn (remember, projection doesn't expand distance), so the sequence M′n still has M as a limit point (it gets arbitrarily close to a sequence that gets arbitrarily close to M). If we can show that all the M′n lie in ugh(Θ)(πpa), then by closure, we'll get that M lies in the same set so we're done.Ok, so we have M′∞i,n∈ugh(Θ)(πi), that project down and mix to make M′n, and

importantly, we crafted them so they're produceable without closure. Thus, they have preimage points M′′∞i,n∈Θ(π¬h∙πi) (that lack nirvana off-h) Project them down to make M′′i,n∈Θ(π¬h∙πpa), and mix them to make a M′′n in the same set (which still lacks nirvana off-h), and this updates to make M′n via Lemma 28, as we'll show shortly.Starting with the M′′∞i,n, we know that update, project, mix equals M′n via going M′∞i,n, M′i,n, M′n. Then, update-project-mix equals project-update-mix equals project-mix-update, which is the path we took. Therefore, all the M′n lie in ugh(Θ)(πpa), which is closed, so M (arbitrary in the closed convex hull of projections) lies in the same set, establishing the reverse subset direction and thus equality,

ugh(Θ)(πpa)=¯¯¯¯¯¯¯¯c.h(⋃π≥πpaprπ,πpa∗(ugh(Θ)(π)))

Part 4: Now that we're halfway done,let's look at the "intersection of preimages of stubs from below" direction of consistency, ugh(Θ)(πpa)⊆⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa)). If we ignore the closure part and work with the raw update set sans closure, we can fix a M in ugh(Θ)(πhipa), take a preimage point in Θ(π¬h∙πhipa), project it down to Θ(π¬h∙πlopa) by consistency, then update it to get exactly the projection of M (again, Lemma 28) Then, when we take the closure, we can just take our M in the closure, fix a sequence in the raw update set sans closure Mn that limits to M, project down, getting M′n in the raw update set ugh(Θ)(πlopa) sans closure, and then the limit point M′ lies in ugh(Θ)(πlopa) by closure, and by continuity of projection, M′ is the projection of M.

Since the sets get bigger as you go down, we can invoke Lemma 6 to swap out the intersection of preimages of all stubs below you, for the intersection of preimages of stubs of the form πnpa, this will be important later.

Now, it's trivial to show that ugh(Θ)(πpa)⊆⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa)) because we've established that projecting down makes a subset, and projection commutes, so any M∈ugh(Θ)(πpa) projects down into ugh(Θ)(πnpa) for all n.

All that's left now is the reverse subset direction,

ugh(Θ)(πpa)⊇⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa))

Sadly, this one will require us splitting into the nirvana-containing (and thus causal) cases and the nirvana-free cases, and it's a really difficult one to show.

Part 5: Let's address the nirvana-free case, we'll use a nifty trick to control the size of the preimage points we select.

Ok, let's say you have a M with some λ1 and b1 value. And you take M′ that's a preimage point, but its λ and b values are just... waaay too high. We want to have a preimage point with reasonable values, in order to apply bounding arguments. What you do, is find a minimal-point Mmin below M′, so M′=Mmin+M∗. Now, what you do, is swap out M∗ ie (m∗,b∗), for (m∗|h,b∗+m∗(0★hg)). This is an sa-measure, because

b∗+m∗(0★hg)+(m∗|h)−(1)=b∗+m∗(0★hg)+m∗−(1★h0)

≥m∗−(0★hg)+m∗−(1★h0)=b∗+m∗−(1★hg)≥b∗+m∗−(1)≥0

Now, consider updating Mmin+(m∗|h,b∗+m∗(0★hg)) instead (it's an a-measure, it has less negative parts than M∗, and is present by nirvana-free upper-completion). This gets you the update of Mmin, plus... (c(m∗|h),b∗+m∗(0★hg)) (remember, 0 measure off-h). Which is the

exact same thingyou'd get by updating M∗, so when we updated our new sum, we hit M exactly.However, this sum is special, because we

canstick some decent bounds on its λ and b value! For starters, its b value is less than b1 (updating only adds on b-mass, and it updates to M). And as for the λ value... well, Mmin has its λ bounded above by λ⊙ (of the original Θ) due to being a minimal point. And in the worst-case, all of the measure in M came from the thing we added, so m∗|h has a measure of λ1 or less. So our bound on the λ value is λ⊙+λ1.Armed with this knowledge, we can begin to prove the last bit of consistency in the nirvana-free case. Take a M in the intersection of preimages. It projects down to make Mn in ugh(Θ)(πnpa). Projection preserves λ and b, so they all have the same λ1,b1 bounds. Because we don't need to close in the nirvana-free case, we get a preimage point of M′n in Θ(π¬h∙πnpa) From our earlier considerations, we can always pick M′n such that its λ is ≤λ⊙+λ1, and its b is ≤b1, although we'll be using a bound of max(b1,b⊙).

Now, we're going to have to be extremely careful here. Let the point M′n,j be defined as:

If j<n, then M′n,j is some arbitrary point in Θ(π¬h∙πnpa), with λ equal to or below λ⊙+λ1, and b equal to or below max(b1,b⊙), which always exists by all minimal points obeying the λ⊙+b⊙ bound.

If j=n, then M′n,j=M′n.

If j>n, then M′n,j=pr(π¬h,πjpa),(π¬h,πnpa)∗(M′j)

Then, the tuple of M′n,j for all n is a point in:

∏n(Θ(π¬h∙πnpa)∩{(λμ,b)|λ≤λ⊙+λ1,b≤max(b1,b⊙)})

Equipped with the product topology. In particular, this is a product of compact sets, so by Tychonoff's theorem, it's compact. Thus, we can get a convergent subsequence of the tuples. On this subsequence, all the M′n,j converge to a limit point M′n,∞, regardless of n.

Also, M′n,∞ projects down to M′m,∞ if n≥m, because for large enough j, the projection of M′n,j will

alwaysbe M′m,j, and by continuity of projection, the projection of M′n,∞ must be M′m,∞Ok, so we've got an infinite sequence of M′n,∞ for all n that all project down onto each other. Another nice feature is that updating M′n,∞ produces Mn. This is because, when j climbs high enough, M′j,j projects down to M′n,j, and M′j,j is just M′j which updates to Mj, which projects down to Mn. By Lemma 28, update-then-project equals project-then-update, so M′n,j must update to Mn, for all sufficiently high j. The preimage of a single point is closed, so past a certain point, the M′n,j are wandering around in the preimage of Mn, so M′n,∞ also updates to Mn.

Now, our next step is, does the M′n,∞ sequence in Θ(π¬h∙πnpa) pick out a single point M′ in Θ(π¬h∙πpa) that projects down accordingly? Yes it does. Just intersect all the preimages of single points, they're nested in each other and compact so the finite intersection property holds, and if the intersection wasn't composed of a single point, you'd have two distinct points with a difference at some finite time, but projecting down to any finite time the two distinct points are identical, so there can only be a single point in the intersection. Further, it must lie in Θ(π¬h∙πpa), because you can project it down to M′n,∞ in Θ(π¬h∙πnpa) for any n, which, by consistency for Θ, you can also project down to Θ((π¬h∙πpa)n) (projecting down further), so it's present in the intersection of all the preimages, certifying that it's in the appropriate set.

Now, finally... does M′, when updated, produce M, certifying that the point in the intersection of preimages is also in the raw update set? Well, let's say it didn't. Then we get a M′′ that's not equal to M, so projecting down to some finite n should suffice to observe that. However, projecting M′′ and M down produces... Mn. This is because of Lemma 28, update-then-project equals project-then-update. Projecting M′ down makes M′n,∞, which updates to Mn.

So, no finite stage suffices to observe the difference between the updated form of M′ and M itself, so they must be identical, certifying ugh(Θ)(πpa)⊇⋂n(prπpa,πnpa∗)−1(ugh(Θ)(πnpa)) for the nirvana-free case.

Part 6: Let's move to the nirvana-case where we can leverage causality. We'll be showing this in a rather nonstandard way. We're going to pick a π≥πpa, and show that our M of interest in the intersection of preimages can be written as a limit of points projected down from ugh(Θ)(π), establishing that M lies in the closed convex hull of points from above, which we've already shown equals ugh(Θ)(πpa).

Ok, so M is in the intersection of preimages. Project it down to all the ugh(Θ)(πnpa), getting a batch of points Mn from them. This is the raw update set, so within 2−n or less distance from Mn, there's a M′n in the raw update sans closure, which has a preimage point M′′n that lies in Θ(π¬h∙πnpa).

Now, pick some arbitrary policy above π¬h∙πpa, which can be written as π¬h∙π. Moving on even further, by causality, we can get a point M′′∞n∈Θ(π¬h∙π) that projects down to M′′n. Update M′′∞n to get a M′∞n∈ugh(Θ)(π), which then (by our earlier thing about how a set equaled the closed convex hull of projections down from above), projects down to a M′hin∈ugh(Θ)(πpa).

Now, we can ask whether the sequence M′hin limits to M itself. ugh(Θ)(πpa) is closed, so this would certify that M lies in the appropriate set.

First, observe that the projection of M′hin down to πnpa is M′n. This is by Lemma 28, update-then-project equals project-then-update. M′′∞n projects down to M′′n, which updates to M′n, so M′n must be what you get by updating M′′∞n to M′∞n, and projecting down to πpa (making M′hin), and projecting that down to πnpa.

Now, because projection preserves the b term, and M′hin projects down to M′n which is within 2−n of Mn (not much of a difference in the b terms), and Mn has the same b term as M, we can certify convergence of the b term at least. Now for convergence of the measure components. Again, M′hin projects down to M′n which is within 2−n of Mn (not much difference before timestep n, shrinking increasingly low), and Mn perfectly mimics what M does before timestep n. So, M′hin behaves increasingly closely to M for everything before time n, which increases without bound. Increasingly close matches on increasingly large initial segments of what happens mean that M′hin must limit to M itself, certifying that M lies in ugh(Θ)(πpa) for the causal cases.

That's the last bit we needed! We're finally done with consistency now. This just leaves the hausdorff-condition and the extreme-point condition and pseudocausality and causality.

Condition 9:Hausdorff-continuity:What we need to do for our setup to even approach this is to show that updating the preimage of the nirvana-free part of Θ(π¬h∙πpa), produces

exactlythe preimage of the nirvana-free part of ugh(Θ)(πpa).One direction, we can get easily. If you fix a M′∞ in the preimage of the nirvana-free part of Θ(π¬h∙πpa), it projects down to a M′∈Θ(π¬h∙πpa)∩NF, that updates to a M∈ugh(Θ)(πpa), then by Lemma 28, project-then-update equals update-then-project, so M′∞ must update to a M∞ that projects down to M, certifying that updating the preimage of the nirvana-free part of Θ(π¬h∙πpa) produces a subset of the preimage of the nirvana-free part of ugh(Θ)(πpa).

In the other direction, fix a M∞ in the preimage of the nirvana-free part of ugh(Θ)(πpa). It projects down to a M in ugh(Θ)(πpa)∩NF, and by Lemma 27, M wasn't introduced in the closure, so it has a preimage point M′∈Θ(π¬h∙πpa)∩NF.

Now, how do we extend M′ to craft a M′∞ that updates to M∞? Well, we can split into two parts. What happens on-h, and what happens off-h? For the off-h part, the post-update part has everything folded into the b term, while the pre-update part has an actual measure specified

everywhere. Thus, our M′∞ should have the same off-h part as M′ to project down accordingly, so updating folds it into the same b term as M∞ has.Now, for the on-h part, it's a bit more complicated. M∞ specified what happens for all infinite histories with h as a prefix. However, M and M′ only specify

partof that data, but fortunately agree on that part. Thus, for M′∞, you can just extend with the conditional probabilities of M∞, to perfectly mimic it on-h. This makes a M′∞ in the preimage that updates to M∞.Ok, so the appropriate preimages for Hausdorff-continuity (post-update) are made exactly by updating the preimages for Hausdorff-continuity (pre-update). Now, updating is a continuous linear operator. We're mapping from the Banach space M±((A×O)ω)⊕R to the Banach space M±(h(A×O)ω)⊕R.

Well, this isn't quite right, your actions and observations may vary depending on where you are in history, but the general thing of "restrict to signed measures over infinite histories with h as a prefix" still checks out. Updating is still a continuous linear operator between Banach spaces, by Lemma 8 of Section 1.

Also, all continuous linear operators between Banach spaces are bounded, and thus Lipschitz-continuous at 0, and thus Lipschitz-continuous everywhere due to linearity. So, when we push two points that are only ϵ apart through the update, they're now Cϵ apart at most, where C is a finite constant.

We're going to have a lot of points. Unusually enough, we'll be using the standard formulation of Hausdorff-continuity for our original Θ, that for all ϵ, there's a δ where two partial policies πpa and π′pa that are δ or less apart have (pr∞,πpa∗)−1(Θ(πpa)∩NF∩{≤⊙}) (and the analogous set for π′pa) being only ϵ apart in Hausdorff-distance.

Fixing your ϵ, you're gonna want δ to be low enough to force a ϵC difference between the clipped preimages, and δ<ϵC. It's

highlyadvised to sketch out how our points interact and what sets they're in. A superscript of infinity will be used to denote points in the preimages of the ugh(Θ)(πpa) sets (or Θ(π¬h∙πpa)) (ie, at the infinite levels), and a superscript of "u" specifies post-update while its lack is pre-update.Anyways, here's our points.

Mu,∞ lies in the preimage of ugh(Θ)(πpa)∩NF, and it's our point that we want to find a point nearby. λ will refer to the λ value of this thing.

Projecting Mu,∞ down to ugh(Θ)(πpa)∩NF makes Mu.

We can find a minimal point below Mu, Mu,min in ugh(Θ)(πpa)∩NF. Mu,min+Mu,∗=Mu.

A nirvana-free point wasn't introduced by the closure, and it has a minimal point in its preimage, so there's a Mmin in Θ(πpa) that updates to Mu,min, and respects the λ⊙+b⊙ bound of Θ.

Let Mlo be defined as Mmin+((mu,∗)−,−(mu,∗)−(1)). We're extending the negative-measure part of Mu,∗ back to its original domain by sticking an h prefix on everything, and saying it's 0 everywhere else. this is an a-measure that lies in Θ(π¬h∙πpa)∩NF∩{≤⊙} (because Mmin respects the λ⊙+b⊙ bound, and the thing that we added has a λ+b value of 0)

Let M be defined as Mlo+((mu,∗)+,bu,∗+(mu,∗)−(1)), it also lies in the same set Updating M makes Mu, because, unpacking M, it's Mmin+Mu,∗, which updates to Mu,min+Mu,∗ which adds up to make Mu.

Our goal now is to explicitly construct a M∞ and Mlo,∞ in the preimage of Θ(π¬h∙πpa)∩NF s.t. they project down onto M and Mlo, Mlo,∞ lies below M∞, and M∞ updates to Mu,∞.

A sufficient way to do this is to make Mlo,∞ and M∞ by, after h, extending the measures further with the conditional probabilities of the measure component of Mu,∞. Extending ((mu,∗)+,bu,∗+(mu,∗)−(1)) with the conditional probabilities of Mu,∞ witnesses that Mlo,∞ lies below M∞. They obviously project down onto M and Mlo.

As for M∞ updating to Mu,∞, the b term and the fragment of the measure that doesn't get ignored by projection down matches because M∞ projects to M which updates to Mu which is the projection of Mu,∞. And, for the fragment of the measure that isn't defined in Θ(π¬h∙πpa), but that must be present on the infinite levels, we copied the conditional probabilities of the measure component Mu,∞, so we've got a match there.

Taking a break from setting up all our damn points for a brief recap, we have a Mlo,∞ that lies in the preimage of Θ(π¬h∙πpa)∩NF∩{≤⊙}, and a M∞ that lies above it (in the preimage of Θ(π¬h∙πpa)∩NF), and it updates to hit Mu,∞ (our original point in the preimage of ugh(Θ)(πpa)∩NF). Now, we can proceed.

So... Mlo,∞ lies in the preimage of Θ(π¬h∙πpa)∩NF∩{≤⊙}. By hausdorff-continuity for Θ and the distance between (π¬h∙πpa) and (π¬h∙π′pa) being below δ because the distance between πpa and π′pa is below δ, and using our earlier thing about how a δ distance means a ϵC difference between the clipped preimages, we can find a point (Mlo,∞)′ in the preimage of Θ(π¬h∙π′pa)∩NF∩{≤⊙} that's that close to Mlo,∞. To go up from Mlo,∞ to M∞ requires adding ((mu,∗)+,bu,∗+(mu,∗)−(1)) (with the measure component extended with the conditional probabilities of the measure component of Mu,∞, obviously).

Also, because the λ value of Mu,∞ is the λ value of Mu, which was made by adding Mu,∗ to an a-measure, an upper bound on the λ value of that a-measure we added onto Mlo,∞ is... λ. Corresponding to the extreme case where all the measure of Mu came from Mu,∗.

Now, we can craft a point (M∞)′ which lies in the preimage of Θ(π¬h∙π′pa)∩NF that's only ϵC+δλ away from M∞. Why? Well, we can start with (Mlo,∞)′, which is only ϵC away from Mlo,∞, and take that positive-measure-thingy we added, and reshuffle the measure on it. With earthmover distance, the δ distance between (π¬h∙π′pa) and (π¬h∙πpa) corresponds to a time-threshold where they start to differ at logγ(δ), and you're moving dirt a γlogγ(δ)=δ difference to account for having to land in the right preimage, and you've got λ at most dirt to move. Then, you just add (Mlo,∞)′ and your reshuffled measure, to get your point (M∞)′. Which is the sum of two components that only differ by ϵC and δλ from the components which sum to make M∞.

Ok, so we have a point M∞ in the preimage of Θ(π¬h∙πpa)∩NF, which updates to Mu,∞ that lies in the preimage of ugh(Θ)(πpa). And a point (M∞)′ in the preimage of Θ(π¬h∙π′pa)∩NF which is (taking into account that δ<ϵC) only ϵC(1+λ) distance away from M∞.

And now we can finish up, because the preimage of ugh(Θ)(π′pa)∩NF is the update of the preimage of Θ(π¬h∙π′pa)∩NF. So, we just update (M∞)′ to get a point (Mu,∞)′ in the preimage of ugh(Θ)(π′pa). And further, the distance between M∞ and (M∞)′ is only ϵC(1+λ) at most. M∞ updates to Mu,∞, and (M∞)′ updates to (Mu,∞)′. And we know that ugh has a Lipschitz constant of C (by being a continuous linear operator between Banach spaces), so Mu,∞ only has a distance of ϵ(1+λ) from a point in the preimage of ugh(Θ)(π′pa).

So, we get Hausdorff-continuity (the Lemma 15 variant).

Condition 8:Extreme Point Condition:We had to defer this because π¬h∙πst

isn'ta stub, so we can't use the extreme point condition we had, and instead must regenerate it completely from scratch.Our first step in this is showing ugh(Θ)(πst)∩NF=¯¯¯¯¯¯¯¯c.h(⋃π≥πstprπ,πst(ugh(Θ)(π)∩NF))

One subset direction is easy, the closed convex hull of projections of nirvana-free stuff must all be in ugh(Θ)(πst) by consistency which we've shown, and all must be nirvana-free. Now for the reverse direction. Let M∈ugh(Θ)(πst)∩NF By Lemma 27, this point wasn't added in the closure, so it has a preimage point M′∈Θ(π¬h∙πst)∩NF. Using all our nice conditions for Θ, we can invoke Lemma 21 to get that M′∈¯¯¯¯¯¯¯¯c.h(⋃π≥(π¬h∙πst)prπ,(π¬h∙πst)(Θ(π)∩NF)), so we can fix a sequence M′n limiting to M where each M′n shatters into M′i,n that came from some M′∞i,n that's nirvana-free and lies in the associated set of a full policy above π¬h∙πst.

Updating the M′n produces a sequence Mn which is nirvana-free, in ugh(Θ)(πst), and limits to M by continuity.

Updating the M′∞i,n into M∞i,n which lie in ugh(Θ)(πi)∩NF, projecting down to get Mi,n, and mixing them, produces Mn, by our usual Lemma 28 argument.

This witnesses that all the Mn lie in c.h(⋃π>πstprπ,πst(ugh(Θ)(π)∩NF))

Thus, M lies in the closed convex hull of projections of nirvana-free stuff from above. What do we do with this? Well, now we can invoke Lemma 20, since we have Hausdorff-continuity proved, to conclude that c.h(⋃π≥πstprπ,πst(ugh(Θ)(π)∩NF)) is closed, so we didn't

reallyneed the closed convex hull (which we've already shown is the same as ugh(Θ)(πst)∩NF)And we now know that ugh(Θ)(πst)∩NF=c.h(⋃π≥πstprπ,πst(ugh(Θ)(π)∩NF))

Now, we can take a minimal extreme nirvana-free point Mex in ugh(Θ)(πst). It must be minimal and extreme and nirvana-free in the original set. If it wasn't minimal in the original set, all minimals below it would be nirvana-free too, witnessing its nonminimiality in the restricted set. And if it wasn't extreme in the original set, then the points that mix to make it must all be nirvana-free too, since it's nirvana-free, so we have a witness of non-extremeness in ugh(Θ)(πst)∩NF.

Ok, so it's extreme and nirvana-free. It must also be extreme in the convex hull set, but, since it can't be produced by mixtures, there's a M∞ in

someugh(Θ)(π)∩NF that projects down to Mex, establishing the extreme point condition.That just leaves causality and pseudocausality.

Condition C:CausalityOk, we pick a πpa and a point in ugh(Θ)(πpa) Can we make an outcome function for everything that includes our point? By our proof of full causality in the first part of the Isomorphism theorem (finite-to-full direction), this can be done as long as all other conditions are met and we can make an outcome function for any point in any ugh(Θ)(πst). So, let's just establish finitary causality. Fix some πst and some M∈ugh(Θ)(πst).

Since M is in the updated set, there's a sequence Mn that limits to M that we don't need closure to get. There's a λ and b bound on this sequence because it converges, call those bounds λ◯ and b◯. Now, we can take a M′n∈Θ(π¬h∙πpa) that updates to Mn. We can use causality for Θ to get an outcome function for M′n.

We don't have to worry about nirvana-off-h, because M′n has no nirvana off-h, and the projection of M′n down to Θ(π¬h