The previous proofs are here.
Proposition 5: If , then the condition "there is a where, " is equivalent to "there is a compact s.t. "
Proof sketch: One direction is immediate from the Compactness Lemma. For showing that just a bound on the values suffices to be contained in a compact set, instead of a bound on the and values to invoke the Compactness Lemma, we use a proof by contradiction where we can get a bound on the values of the minimal points from just a bound on the values.
Proof: In one direction, assume there's a compact s.t. , and yet there's no upper-bounding on the values. This is impossible by the Compactness Lemma, since .
In the other direction, assume there's a bound on for the minimal points. Fix some arbitrary for the rest of the proof. Now, we will show that all minimal points have , and letting us invoke the Compactness Lemma to get that everything is in a suitable compact set . The first bound is obvious. Since came from a minimal point, it must have as an upper bound.
For the other one, by contradiction, let's assume that there's a minimal point where . Then, we can write as:
The first component, is our fixed minimal point of interest. The second component is an sa-measure, because , due to the upper bound on the value of minimal points. The third component is also a nonzero sa-measure, because is nonnegative (it came from a minimal point), and by assumption, . Hang on, we wrote a minimal point as another minimal point , plus two sa-measures (one of which is nonzero), so can't be minimal, and we have a contradiction.
Therefore, all have . Now that we have bounds on and for minimal points, we can invoke the Compactness Lemma to conclude that everything is in a compact set.
Proposition 6: only occurs when there's only one minimal point of the form .
Proof: Unpacking the expectations, and in light of Proposition 3,
So, take a minimal a-measure that minimizes . One must exist because we have and bounds, so by the Compactness Lemma, we can restrict our attention to an actual compact set, and continuous functions from a compact set to have a minimum, so there's an actual minimizing minimal point.
must be 0, because otherwise which contradicts . Further, since , said must be the lowest possible amongst minimal points.
So, we have a minimal point of the form where is the lowest possible amongst the minimal points. Any other distinct minimal point must be of the form , where . This other minimal point can be written as , where the latter component is an sa-measure, so it's not minimal. Thus, there's only one minimal a-measure and it's of the form .
Proposition 7: Renormalizing a bounded inframeasure produces a bounded infradistribution, if renormalization doesn't fail.
Proof sketch: Our first order of business is showing that our renormalization process doesn't map anything outside the cone of sa-measures. A variant of this argument establishes that the preimage of a minimal point in must be a minimal point in , which quickly establishes positive-minimals and bounded-minimals for . Then, we verify the other conditions of a bounded infradistribution. Nonemptiness, closure, and convexity are very easy, upper-closure is shown by adding appropriately-scaled sa-measures such that, after renormalization, they hit whatever sa-measure you want. Then, finally, we just have to verify that our renormalization procedure is the right one to use, that it makes and .
Proof: First up, we need to show that after renormalization, nothing gets mapped outside the cone of sa-measures. Observe that the renormalization process is injective. If two points are distinct, after a scale-and-shift, they'll still be distinct.
Let B be our original set and be our renormalized set. Take a point in , given by . Undoing the renormalization, we get .
By decomposition into a minimal point and something else via Theorem 2, we get that
where . Renormalizing back, we get that
, obviously, because is the minimal value amongst the minimal points. So, the first component is an a-measure, the second component is an sa-measure, so adding them is an sa-measure, and then we scale by a nonnegative constant, so is an sa-measure as well.
This general line of argument also establishes positive-minimals and bounded-minimals, as we'll now show. If the isn't 0, then we just wrote as
And the first component lies in , but the latter component is nonzero, witnessing that isn't minimal. So, if is minimal in , then , so it must be the image of a single minimal point by injectivity. Ie, the preimage of a minimal point in is a minimal point in .
Scale-and-shift maps a-measures to a-measures, showing positive-minimals, and the positive scale constant of just scales up the upper bound on the values of the minimal points in , showing bounded-minimals.
For the remaining conditions, nonemptiness, closure, and convexity are trivial. We're taking a nonempty closed convex set and doing a scale-and-shift so it's nonempty closed convex.
Time for upper-completeness. Letting be our original set and be our renormalized set, take a point in . By injectivity, has a single preimage point . Undoing the renormalization by multiplying by (our addition of is paired with to undo the renormalization on that one), consider This lies in by upper-completeness, and renormalizing it back produces , which is in , so is upper-complete.
That just leaves showing that after renormalizing, we're normalized.
For the other part,
And we're done.
Lemma 6: is a continuous linear operator.
Proof sketch: First show linearity, then continuity, for the operator that just maps a signed measure through , using some equation-crunching and characterizations of continuity. Then, since is just the pair of that and the identity function, it's trivial to show that it's linear and continuous.
We'll use to refer to the function defined by , where is a measurable subset of and . Ie, this specifies what the measure is in terms of telling you what value it assigns to all measurable subsets of .
We'll use to refer to the function given by .
Our first order of business is establishing the linearity of . Observe that, for all measurable , and being real numbers, and being signed measures over ,
So, and we have linearity of .
Now for continuity of . Let limit to . The sequence converging to in our metric on is equivalent to:
So, if fails to converge to , then there is some continuous function that witnesses the failure of convergence. But, because is a continuous function , then , and also , so:
The key step in the middle is that limits to , so limits to , by our characterization of continuity. Thus, we get a contradiction, our that witnesses the failure of convergence actually does converge. Therefore, limits to if limits to , so is continuous.
To finish up, continuity for comes from the product of two continuous functions being continuous ( which we showed already, and because duh), and linearity comes from:
Proposition 8: If and is a continuous function , then
Proposition 9: is a (bounded) inframeasure if is, and it doesn't require upper completion if is surjective.
Proof sketch: Nonemptiness is obvious, and showing that it maps sa-measures to sa-measures is also pretty easy. Closure takes a rather long argument that the image of any closed subset of sa-measures over , through , is closed, which is fairly tedious. We may or may not invoke upper completion afterwards, but if we do, we can just appeal to the lemma that the upper completion of a closed set is closed. Convexity is immediate from linearity of .
For upper completion, we can just go "we took the upper completion" if isn't surjective, but we also need to show that we don't need to take the upper completion if is surjective, which requires crafting a measurable inverse function to g via the Kuratowski-Ryll-Nardzewski selection theorem, in order to craft suitable preimage points.
Then we can use LF-Duality to characterize the induced function, along with Proposition 8, which lets us get positive-minimals, bounded-minimals, and normalization fairly easily, wrapping up the proof.
Proof: Nonemptiness is obvious. For showing that it takes sa-measures to sa-measures, take an , and map it through to get . is an sa-measure, so . Now, we can use Lemma 5 to get:
So the term is indeed big enough that the image of is an sa-measure.
For closure, fix a sequence of limiting to some , with preimage points . Due to convergence of there must be some bound on the . preserves those values, so is an upper bound on the . Since the are sa-measures, is a lower bound on the values. Since converges to , converges to , so there's a upper bound on the values. Further,
So, for all n, , so we have an upper bound on the and values. Now we can invoke the Compactness Lemma to conclude that there's a convergent subsequence of the , with a limit point , which must be in since is closed. By continuity of from Lemma 6, must equal , witnessing that . So, is closed. Now, if we take upper completion afterwards, we can just invoke Lemma 2 to conclude that the upper completion of a closed set of sa-measures is closed.
Also, is linear from Lemma 6, so it maps convex sets to convex sets getting convexity.
Now for upper completion. Upper completion is immediate if isn't surjective, because we had to take the upper completion there. Showing we don't need upper completion if is surjective is trickier. We must show that is a surjection from to .
First, we'll show that where is an open subset of is a measurable subset of . In metrizable spaces (of which is one), every open set is a set, ie, it can be written as a countable union of closed sets. Because our space is compact, all those closed sets are compact. And the continuous image of a compact set is a compact set, ie closed. Therefore, is a countable union of closed sets, ie, measurable.
is a Polish space (all compact metric spaces are Polish), it has the Borel -algebra, and we'll use the function . Note that is closed and nonempty for all due to being a continuous surjection. Further, the set equals for all open sets . In one direction, if the point is in the first set, then there's some point where . In the other direction, if a point is in , then there's some point where so is nonempty.
Thus, is weakly measurable, because for all open sets of , and is measurable. Now, by the Kuratowski-Ryll-Nardzewski Measurable Selection Theorem, we get a measurable function from to where so , and is an injection.
So, we can push any sa-measure of interest through (which preserves the amount of negative measure due to being an injection), to get an sa-measure that, when pushed through recovers exactly. Thus, if , and you want to show , just consider
So, since due to upper-completeness, then And we have shown upper-completeness of if is a surjection.
We should specify something about using LF-Duality here. If you look back through the proof of Theorem 5 carefully, the only conditions you really need for isomorphism are (on the set side) being closed, convex, and upper complete (in order to use Proposition 2 to rewrite appropriately for the subsequent arguments, we have these properties), and (on the functional side), being concave (free), if (by proof of Theorem 4, comes from upper completeness), and continuous over (showable by Proposition 8 that , and the latter being continuous since is an infradistribution)
It's a bit of a pain to run through this argument over and over again, so we just need to remember that if you can show closure, convexity, upper completeness, and the expectations to be continuous, that's enough to invoke LF-Duality and clean up the minimal point conditions. We did that, so we can invoke LF-Duality now.
Time for normalization. From Proposition 8, the function we get from is uniquely characterized as: . So,
and normalization is taken care of.
For bounded-minimals/weak-bounded-minimals, since is the LF-dual of , we can appeal to Theorem 5 and just check whether is Lipschitz/uniformly continuous. if , then according to the sup metric on and , respectively, which (depending on whether we're dealing with Lipschitzness or uniform continuity), implies that , or for uniform continuity. So, we get: (or for uniform continuity), thus establishing that and being sufficiently close means that doesn't change much, which, by Theorem 5, implies bounded-minimals/weak-bounded-minimals in .
For positive-minimals it's another Theorem 5 argument. If , then , so: And we have monotonicity for , which, by Theorem 5, translates into positive-minimals on .
Lemma 7: If , then for all decompositions of into ,
This is easy. Decompose into . To derive a contradiction, assume there exists a nonminimal that decomposes into where . Then,
Thus, we have decomposed our minimal point into another point which is also present in , and a nonzero sa-measure because there's a nonzero so our original "minimal point" is nonminimal. Therefore, all decompositions of a minimal point in the mixture set must have every component part being minimal as well.
Proposition 11: A mixture of infradistributions is an infradistribution. If it's a mixture of bounded infradistributions with Lipschitz constants on their associated functions of , and , then the mixture is a bounded infradistribution.
Proof sketch: Nonemptiness, convexity, upper completion, and normalization are pretty easy to show. Closure is a nightmare.
The proof sketch of Closure is: Take a sequence limiting to . Since each approximating point is a mixture of points from the , we can shatter each of these into countably many . This defines a sequence in each (not necessarily convergent). Then, we take some bounds on the and manage to translate them into (rather weak) i-dependent bounds on the sequence. This lets us invoke the Compactness Lemma and view everything as wandering around in a compact set, regardless of . Then, we take the product of these compact sets to view everything as a single sequence in the product of compact sets, which is compact by Tychonoff's theorem. This is only a countable product of compact metric spaces, so we don't need full axiom of choice. Anyways, we isolate a convergent subsequence in there, which makes a convergent subsequence in each of the . And then, we can ask "what happens when we mix the limit points in the according to ?" Well, what we can do is just take a partial sum of the mixture of limit points, like the i from 0 to 1 zillion. We can establish that gets arbitrarily close to the upper completion of a partial sum of the mixture of limit points, so lies above all the partial sums of our limit points. We show that the partial sums don't have multiple limits, then, we just do one more invocation of Lemma 3 to conclude that the mixture of limit points lies below . Finally, we appeal to upper completion to conclude that is in our mixed set of interest. Whew!
Once those first 4 are out of the way, we can then invoke Theorem 5 to translate to the view, and mop up the remaining minimal-point conditions.
First, nonemptiness. By Theorem 5, we can go "hm, the are monotone on , and everywhere else, and , so the affine functional lies above the graph of ". This translates to the point being present in all the . Then, we can just go: , so we have a point in our set.
For normalization, appeal to Proposition 10 and normalization for all the . and .
Convexity is another easy one. Take a . They shatter into . Then, we can just go:
and then, by convexity of the , , so we wrote as a mixture of points in .
Upper completion is another easy one, because, if , then you can go
And by upper completion.
That leaves the nightmare of closure. Fix a sequence limiting to . You can think of the as . We can shatter the into , where can be thought of as .
Now, since converge to something, there must be an upper bound on the and terms of the sequence, call those and . Now, for all n and all , so, for all n and i, .
Also, for all n and , and reshuffling, we get which then makes . Further, due to being a sa-measure, , so for all n and i, .
Ok, so taking stock of what we've shown so far, it's that for all i, the sequence is roaming about within And, by the Compactness Lemma, this set is compact, since it's got bounds (weak bounds, but bounds nonetheless). Defining
where , we can view everything as one single sequence wandering around in the product of compact sets. By Tychonoff's theorem (we've only got a countable product of compact metric spaces, so we don't need full axiom of choice, dependent choice suffices), we can fix a convergent subsequence of this, and the projections of this subsequence to every converge.
Ok, so we've got a subsequence of n where, regardless of i, converge to some (by closure of ). How does that help us? We don't even know if mixing these limit points converges to something or runs off to infinity. Well... fix any j you like, we'll just look at the partial sum of the first j components. Also fix any you please. On our subsequence of interest, the converge to , and in all i, the converge to . So, let n be large enough (and in our subsequence) that , and , we can always find such an n.
Now, is a well-defined point (because it's a finite sum of points plus a convergent sequence as witnessed by the well-definedness of which breaks down as ) It also lies in the upper completion of the single point . We'll show that this point is close to . Since we're working in a space with a norm,
This will come in handy in the later equations.
So, is less than away from the upper completion of the point , which is a closed set (Minkowski sum of a closed and compact set is closed). can be shrank to 0 with increasing n, so has distance 0 from the upper completion of said partial sum, and thus lies above the partial sum!
Abbreviating as , we get that all the lie in , and are all sa-measures. Thus, if the sequence converges to a unique point, then said limit point is , and all the , so would lie in . Further, by Lemma 3, , since that set is compact, so lies above , and would lie in by upper-completeness.
So, all that's left to wrap up our closure argument is showing that the sequence has a single limit point. Since it's wandering around in which is compact by Lemma 3, there are convergent subsequences. All we have to show now is that all convergent subsequences must have the same limit point.
Assume this is false, and there's two distinct limit points of the sequence , call them and . Because it's impossible for two points to both be above another (in the minimal-point/adding-points sense), without both points being identical, either , or vice-versa. Without loss of generality, assume . Since the latter is a closed set, must be away for some . Fix some j from the subsequence that is a limit point of, where . There must be some strictly greater from the subsequence that is a limit point of.
Further, the are nonzero. Also, no can be the 0 point, because , and if , then , which is impossible by normalization. So, lies strictly below . Also, lies below , because for all the ,
so for all . The sequence that limits to is roaming around in this set, which is closed because the sum of a compact set (a single point) and a closed set is closed. So, lies above which lies above . Thus, . However, is or less distance from , which must be distance from , and we have a contradiction.
Ok, so the sequence of partial sums has a single limit point, which is , and all the , so , and