LBIT Proofs 2: Propositions 10-18

post by Diffractor · 2020-12-16T03:45:42.643Z · LW · GW · 0 comments

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Proposition 10: Mixture, updating, and continuous pushforward preserve the properties indicated by the diagram, and always produce an infradistribution.

We'll start with showing that mixture, updating, and continuous pushfoward are always infradistributions, and then turn to property verification.

We know from the last post that mixture, updating, and continuous pushfoward preserve all infradistribution properties (although you need to be careful about whether mixture preserves Lipschitzness, you need that the expected value of the Lipschitz constant is finite), but we added the new one about compact almost-support, so that's the only part we need to re-verify.

To show that mixture has compact almost-support, remember that 

Now, fix an , we will craft a compact set that accounts for all but  of why functions have the expectation values they do. There is some n where , where  is the Lipschitz constant of the infradistribution . Then, let  be , the union of the compact -almost-supports for the infradistributions . This is a finite union of compact sets, so it's compact.

Now we can go:



The first equality is reexpressing mixtures, and the first inequality is moving the expectation outside the absolute value which doesn't decrease value, then we break up the expectation for the second equality. The second inequality is because the gap between  and  has a trivial upper bound from the Lipschitzness of , and for , we have that  and  agree on the union of the -almost-supports for the , so a particular infradistribution, by the definition of an almost-support, has these two expectations having not-very-different values. Then we just pull the gap between  and  out, and use the fact that for the mixture to work, , and we picked n big enough for that last tail of the infinite sum to be small. Then we're done.

Now, we will show compact almost-support for  assuming  has compact almost-support. Fix an . Your relevant set for  will be 

Where the first term is a compact set that is a -almost-support for , and that last set is a sort of "this point must be likely enough".  will be the Lipschitz constant of the original . Yes, this intersection may be empty.

Now, here's how things go. Let  and  agree on that intersection. (if it's the empty set, then it can be any two functions). We can go:



So far, this is just a standard sequence of rewrites. The definition of the update, pulling the fraction out, using  to abbreviate the rescaling term, and unpacking what  means.

Now, let's see how different  and  are on the set . One of two things will occur. Our first possibility is that an  in that compact set also has . Then

and  were selected to be equal on that set, so the two functions will be identical on that point. Our second possibility is that  in that compact set will have . In that case,


Because .

Putting this together,  and  are only  apart when restricted to the compact set . By Lemma 2, we can then show that

And, we also know that:

Because . Making that substitution, we have:



Backing up to earlier, we had established that

and from shortly above, we established that

Putting these together,

For any two functions  and  which agree on 

Witnessing that said set is an -almost-support for .

All we need to finish up is to show that this is a compact set in  equipped with the subspace topology. This can be done by observing that in the original space  it's a compact set, due to being the intersection of a compact set and a closed set. In the subspace topology, if we try to make an open cover of it, all the open sets that cover it in the subspace topology are the restrictions of open sets in the original topology, so we have an open cover of this set in the original topology, and we can make a finite subcover, so it's compact in the subspace topology as well.

Thus, for any , we can make a compact (in -almost-support for , so  has compact almost-support and we've verified the last condition for an update of an infradistribution to be an update.

Now for deterministic pushfoward. Fix an , and let your appropriate set for  be  where  is a compact -almost-support for . The image of a compact set is compact, so that part is taken care of. We still need to check that it's an -almost-support for . Let  be equal on this set. Then


And we're done. This is because, for any point , feeding it through  makes a point in , and feeding it through  and  produces identical results because they agree on . Therefore,  and  agree on  and thus can have values only  apart, which is actually upper-bounded by  is thus a compact -almost-support for , and this can be done for any , so  has compact almost-support.

Since these three operations always produce infradistributions (as we've shown, we verified the last condition). Updating only has two properties to check, preserving homogenity when  and cohomogenity when , so let's get that knocked out.

Homogenity using homogenity for h


Cohomogenity using cohomogenity for h








Now for mixtures, we'll verify homogenity, 1-Lipschitzness, cohomogenity, C-additivity, and crispness.

Homogenity:

1-Lipschitz:


Cohomogenity:


C-additivity:

Crispness: Observe that both homogenity and C-additivity are preserved, and crispness is equivalent to the conjunction of the two.

Now for deterministic pushforwards, we'll verify homogenity, 1-Lipschitzness, cohomogenity, C-additivity, crispness, and sharpness.

Homogenity:

1-Lipschitzness:

Cohomogenity:


C-additivity:

Crispness: Both homogenity and C-additivity are preserved, so crispness is preserved too.

Sharpness:

And  is the image of a compact set, so it's compact. And we're done!

Proposition 11: The inf of two infradistributions is always an infradistribution, and inf preserves the infradistribution properties indicated by the diagram at the start of this section.

We'll first verify the infradistribution properties of the inf, and then show it preserves the indicated properties if both components have them.

We must check monotonicity, concavity, normalization, Lipschitzness, and compact almost-support. For monotonicity, if , then

This was done by monotonicity for the components. For concavity,





The first  happened because  and  are concave, the second is because .

For normalization, 

And the same argument applies to 0, so the inf is normalized.

For Lipschitzness, the inf of two Lipschitz functions is Lipschitz.

That just leaves compact almost-support. Fix an arbitary , and get a  compact -almost-support for , and a  for . We will show that  is a compact -almost-support for . It's compact because it's a finite union of compact sets.

Now, let  and  agree on . We can go:

There are four possible cases for evaluating this quantity. In case 1,  and . Then our above term turns into . However, since  and  agree on , they must agree on , and only have expectations  apart. Case 2 where  and  is symmetric and can be disposed of by a nearly identical argument, we just do it with  and .

Case 3 where  and  takes a slightly fancier argument. We can go:

The end inequalities are because  and  agree on the -almost-supports of  and , respectively, from agreeing on the union. The two inner inequalities are derived from the assumed inequalities in Case 3.
Thus,

Case 4 where the assumed starting inequalities go in the other direction is symmetric. So, no matter which infradistributions are lower in the two infs, we have

And we're done, we made a compact almost-support for  assuming an arbitrary . So the inf of two infradistributions is a infradistribution.

Now to verify homogenity, 1-Lipschitzness, cohomogenity, C-additivity, crispness, and sharpness preservation.

Homogenity:


1-Lipschitzness:

Now we can split into four cases. In cases 1 and 2 where the infs turn into  (and same for  in case 2), we have:

(and same for ), and we're done with those cases. In cases 3 and 4 where the infs turn into  (and vice-versa for case 4), we have:

Because  and  are 1-Lipschitz. Thus,

A symmetric argument works for case 4. So, no matter what,

And we're done, the inf is 1-Lipschitz too.

Cohomogenity:




C-additivity:

Crispness: Homogenity and C-additivity are both preserved, so crispness is preserved.

Sharpness:

And we're done.

Proposition 12: 





Proposition 13: If a family of infradistributions  has a shared upper bound on the Lipschitz constant, and for all , there is a compact set  that is an -almost support for all , then , defined as , is an infradistribution. Further, for all conditions listed in the table, if all the  fulfill them, then  fulfills the same property.

We'll first verify the infradistribution properties of the infinite inf, and then show it preserves the indicated properties if all components have them.

We must check monotonicity, concavity, normalization, Lipschitzness, and compact almost-support. For monotonicity, if , then

This was done by monotonicity for all components. For concavity,



The first  happened because  and  are concave, the second is because .

For normalization, 

And the same argument applies to 0, so the inf is normalized.

For Lipschitzness, let  be your uniform upper bound on the Lipschitz constants of the . Then,

And then, for all the , they only think those functions differ by  or less, and the same property applies to the inf by picking a  and  that very very nearly attain the two minimums, and showing that if the infinimums were  apart, you could have  appreciably undershoot , and in fact, undershoot , which is impossible. Thus,

And we're done.

That just leaves compact almost-support. Fix an arbitary . We know there is some  that is a compact -almost-support for all the . We will show that  is an -almost-support for .

Let  and  agree on . We can go:

Pick a  and  that very very very nearly attain the inf. Then we can approximately reexpress this quantity as:

We're approximately in a case where  and , so we can go:

The end inequalities are because  and  agree on the -almost-support of  and . The two inner inequalities are derived from the assumed inequalities in our case. Thus,

And we're done, we made a compact almost-support for  assuming an arbitrary . So the inf of this family of infradistributions is a infradistribution.

Now to verify homogenity, 1-Lipschitzness, cohomogenity, C-additivity, crispness, and sharpness preservation.

Homogenity:


1-Lipschitzness: Same as the Lipschitz argument, everyone has a Lipschitz constant of 1, so the inf has the same Lipschitz constant.

Cohomogenity:




C-additivity:

Crispness: Homogenity and C-additivity are both preserved, so crispness is preserved.

Sharpness:

We do have to check whether or not  is compact, however. We'll start by showing that for an arbitrary , any compact set  where  can't be an -support of  for any . The proof proceeds as follows:

Let  be some point in  but not in . It must be some finite distance away from . Craft a continuous function  supported on  is 1 on  and 0 on . Use the Tietze extension theorem to extend  to all of . Then

However,  and 1 agree on , so  can't be an -almost-support for any .

Thus, in order for there to be a compact set  that's an -almost-support for all , it must be that . Then

because all the  are in it and  is closed. So, the closure of our union is a closed subset of a compact set and thus is compact, so  is minimizing over a compact set and thus is crisp.

Proposition 14: If  and , then the supremum is an infradistribution.

The supremum is defined as:

We'll verify the infradistribution properties of the sup.

We must check monotonicity, concavity, normalization, Lipschitzness, and compact almost-support. For monotonicity, if , then


This was done by  so there's more options available. For concavity,



Pick your  that very very very nearly attain the supremum.





Also, we can verify that:




Therefore, it is a suitable parameter and pair of functions to lower bound 
. Accordingly



Putting all this together, and picking better and better approximations to the two suprema, we can conclude that:

And we have concavity.

For normalization, we're assuming it holds at the start.

Lipschitzness takes a slightly more involved argument. Pick two functions  and , and without loss of generality, assume . Now, what we can do is pick a  and  which approximately obtain the defining supremum for , so we have:

Now, we can note two things. First, 

Therefore, the same , and , and  are suitable things to lower-bound the value of . In particular, we have:


Also, we have the result that:





Because of Lipschitzness of  and . Now we can begin showing our inequalities. So, we've shown that:

Therefore,

With this result, we can go:


Let's save this result for a bit later.
Also, we had:

And we also picked  and  and  to approximately attain the supremum, so we know:

Therefore, we approximately have:

Reshuffling this around a bit, we have:

Using this with our saved result, we can get:



That last inequality was because we assumed at the start without loss of generality that  got an equal or higher expectation than .
Therefore, we have our result that, in general,

And thus, the supremum of two Lipschitz infradistributions is Lipschitz. That just leaves compact almost-support, which is quite tricky to show.

Fix an arbitary , and get a  compact -almost-support for , and a  for . We will show that  is a compact -almost-support for . It's compact because it's a finite union of compact sets.

Now, let  and  agree on . Without loss of generality, assume that  (if not, flip  and ). We'll show that they have similar expectations by showing that  is below a small number (we already know that it's above 0 by our without-loss-of-generality assumption).

We can go:

Where we picked a particular  spectacularly close to the highest possible value s.t. . In particular, if  is 0 or 1, we can ensure that  or  is  itself, by monotonicity of  or  respectively.

For successive arguments, we need  so we have to address those endpoints. Assume . Then, . Then, we have:



The way this works is our substitution, and then using that  and  are identical on , and so are identical on , which is -almost-support of , we can upper-bound  with . And then, we just use that . If , the exact same argument works, just with  and  instead. That leaves the case where , which requires far more involved arguments.

As a recap, we're assuming that , and that , and . Now, we're going to pick out a continuous function with some special properties, so let the set-valued function  be defined as: If , then . Otherwise,  equals the intersection of:

and

We'll find a continuous selection of this set-valued function, so let's start checking the properties needed to invoke the Michael selection theorem. We need that  is paracompact (all polish spaces are paracompact, check), that  is a Banach space (check), that for all  is convex (it's either a single point or the intersection of a rectangle and a half-space, which is convex in both cases), closed (yup, it's either a point or the intersection of two closed sets, ie closed), nonempty, and lower-hemicontinuous.

Nonemptiness isn't too bad to show. It's nonempty for all points in our compact set of interest (the set consisting of a single point), and for x not in said set,  witnesses the nonemptiness, because:


Lower-hemicontinuity is much more challenging to establish. Again, we have a sequence  limiting to , a point , and we must find a subsequence  which limits to .

We can divide into three cases. In the first case,  lies in ,  and infinitely many members of the sequence lie in said set. In particular, since  lies in the compact set, the  pair associated with it must be . Then we can isolate that particular subsequence that lies in the compact set, and have  be , which, by continuity of  and , and the definition of  for  in the compact set, lie in  and limit to  ie .

In preparation for the second and third cases, we'll show that the function  which just takes the second branch of the  function is continuous w.r.t. the Hausdorff-metric. Ie, for all 


is continuous when the space of compact subsets of  is equipped with a Hausdorff distance.

Accordingly, let  limit to . Our task is to show that, no matter how tiny of a number you name, you can find a tail of the  sequence where the Hausdorff distance between  and  is that tiny.

Specifically, we'll show that for all , there is some  where all later  have  within  Hasdorff distance of . Because  and we can shrink  to 0, this shows that the function  is continuous in Hausdorff-distance.

Because  and  and  are continuous functions, there's some very very large  where , and  will only vary by  from that point forward, regardless of which  you pick. Pick some arbitrary . We'll show that it's close to a , and the argument will only depend on distances, not position in sequence, so we can flip it to show the other half of Hausdorff-distance (all points in  are close to a point in ).

We can divide into four possible cases. In cases 1 and 2, we have the following property holding.

With the negation for cases 3 and 4.

And in cases 1 and 3, we have:

With the negation for cases 2 and 4.

In cases 1 and 2, you can let your selected  point be . We have the result that , because:


In order, the first inequality is because  only varies by  over such tiny distances due to continuity of , the second inequality is  being paired with something to be in  so it has a known upper bound on its value, then the third inequality is because , the equality is our definition of our , then for the next inequality using the fact that we're assuming that  has a particular lower bound since we're in cases 1 and 2, Then there's just a cancellation, and  only varying by  over such tiny distances.

You can use nearly identical arguments in cases 1 and 3 to get that, when you define  to be . you have the result that 

Now, in cases 3 and 4, we can let  be: , and then we have:





The first equality is just pair-creation, then the second one is packing up the definition of . The first inequality is because  only varies by  over that distance, the second inequality is because  so it's got the usual lower bound, then the next inequality after that is because we're in cases 3 and 4 so

Then, it's just another " doesn't change much over the tiny distance", moving the 's together, unpacking , and cancelling out. The net result is that we have:

You can use nearly identical arguments in cases 2 and 4 to get that, when you define  to be be  you have the result that .

At this point, we can resume our progress on the four cases and go "ok, in case 1, we have..."


And we know that those properties lead to  being defined as  and  being defined as . And we know that in that case,

So, all we have to check is that  in order to conclude that . Let's do that.


And we have that , accordingly. The first equality was unpacking definitions, then the second was some cancellation, and then the first inequality was because  by assumption so we have . The second inequality was because  doesn't change much over such tiny distances and then it's just trivial cleanup.

Thus, when we picked a point  where  is sufficiently close to , and we're in case 1, we have that there are points , and 


This is from the definitions of  and  in Case 1.

Now, let's address case 2, where


In this case,  is defined as , and  is defined as 
And we know that in that case,

(the first part on the  is the same argument from case 1, the second interval is from the value of )
So, all we have to check is that  in order to conclude that . We know that 

So we can flip this a bit to get

Accordingly, from that, we get:


And we have that , accordingly.The first inequality was definition unpacking and the inequality we just got, then the first equality is just breaking things up a bit, then the second inequality is just observing that , and then  doesn't change much over such tiny distances. 

Thus, when we picked a point  where  is sufficiently close to , and we're in case 2, we have that there are points , and 


This is from the definitions of  and  in Case 2, and the fact that in case 2 we can derive 

Extremely similar arguments to case 2 dispatch case 3 with a resolution of the corresponding  lying in  and


Finally, for case 4, we have:


In this case,  is defined as , and  is defined as 
Trivially, we have:

So, all we have to check is that  in order to conclude that . To do this, we have:
We know that 

So we can flip this a bit to get

Accordingly, from that, we get:


(because )
And we have that , accordingly.

Thus, when we picked a point  where  is sufficiently close to , and we're in case 4, we have that there are points , and 

This is from the definitions of  and  in Case 4, and the fact that in case 4 we can derive  (and same for )

These 4 cases were exhaustive, so we now know that, given any  and sequence of points  limiting to , and any , there is a tail of sufficiently large m's where the distance from any point in  to  is  or less. We can also flip  and  and use our four cases (our argument is symmetric) to show that actually, this is a bound on the Hausdorff distance between  and  was arbitrary, as was the sequence  and the , so this means that  is continuous in Hausdorff-distance.

Ok, we're a bit in the weeds here, how does that help? Well, we were trying to verify the compact almost-support property for the supremum. This requires, as part of it, getting a continuous function with some special properties. We're going to apply a selection function to get it, but we could only take care of the prerequisites that aren't lower-hemicontinuity. And to show lower-Hemicontinuity in general, we needed to take this detour through showing that the modified set-valued function is continuous in Hausdorff-distance. So let's pop back up the stack.

One level back up the stack, we were trying to show lower-Hemicontinuity. It is the property that given any sequence  which limits to , and any , there is some subsequence  and  where  limits to . We dispatched the case where infinitely many elements of the sequence were in our , leaving two cases. There's the case where only finitely elements of that sequence are in that compact set, but the limit point  lies in that set. There's also the case where the limit point  doesn't lie in that set.

Dealing with case 3, we have a sequence  heading to . Strip off all the  that lie in the compact set, making your . And let  be whichever point in  is closest to . Now, by how they were defined, , and  is continuous in Hausdorff-distance, so "take the closest point" is definitely going to get you the convergence you seek to your arbitrarily selected  point.

For case 2, where we're limiting to  from outside the compact set, all we need to show is that  (we don't necessarily have equality because  and  start being different on that compact set), in order to get a sequence  converging to the  point. So, let's do this. Because  lies in , we have that .
The conditions for  to be in  are:

Which is obviously true for , and:

Which is the case because:

By how  was made, and  on that compact set.

Thus, we're done, we verified lower-hemicontinuity for  in all the cases, so we can invoke the Michael selection theorem and get a continuous selection  with three valuable properties. Let's abbreviate  as , for notational convenience. It's projecting it to the first coordinate.  is defined similarly.

Our first notable property is:

(ie, projecting  down to the two coordinates makes functions which perfectly mimic  and  on the compact set of interest)

Our second one is:

And the same for .

And our third notable property is that:

But why do these properties hold of our selection function? Well, when  lies in that compact set, , so our selection function is forced to have its projections mimic  and  on said compact set, taking care of the first one.

For our second property, we have:

Accordingly, we know that the projections to the two coordinates can't be too far away from and  respectively.

For our third property, we have:

Accordingly, the projections to the two coordinates can't mix to exceed the function .

So, where to from here? Well, we have:




Here's why. We assumed at the very start that without loss of generality, we'd take  to be the one with higher expectation value. We found a  that nearly replicated the expectation value of  copies  on a compact almost-support of , namely , and we also have , and similar for  and . And finally, since , that mix must have lower value than . And we're done!  and  were arbitrary except that they agreed on , a compact set, and we got:

Witnessing that said set is a compact -almost-support.  was arbitrary, so  is compactly-almost-supported. This is the last condition needed to check to see that it's an infradistribution.

Proposition 15: All three characterizations of the supremum given in Definition 13 are identical.

So, the first characterization we gave was:

And the second characterization was the least infradistribution greater than  in the information ordering.

And the third characterization was as the concave monotone hull of .

We will use  for these three characterizations of the supremum of two infradistributions and show that they are equal.

Let's begin showing this.

This occurs by monotonicity, any mix of functions which undershoots  must get a lower score because  is an infradistribution.

This is because of convexity of , since it's an infradistribution. The value of the mix is as good or better than the mix of the values.

This is because  (and same for ), so making that swap decreases the value. Also, this quantity is the concave monotone hull of the supremum of . Why? Well,  is our first attempt at assessing the value of a function . However, it isn't necessarily monotone. So,  is the monotone hull, we're saying that if there's a value below you that outscores you, then you should update the value of  to be big enough. And then, to get the concave monotone hull, we replace the lower bound on  with a countable/arbitrary finite mix of functions because any concave function should have the value of the mix be  the mix of the values, so we have to bump the value of  up to at least the mix of the values to not violate concavity. Anyways, now that we know this is , we can go further to:

This is lower because now we're specializing to only certain sorts of probability distributions over , those that are only supported on the first two values, so it's harder to attain suprema. And now,

We swapped out the supremum for a specific term in it in order to do this, and used our given definition of . And then we can specialize  to 1 and  to  itself, to get

Similarly, we could specialize to  and  to get . So taking stock of what we have,

For all functions, so:

(and same for ) We recall that in Proposition 14 we proved that  always makes an infradistribution. Since  is above both component infradistributions, and  was defined as the least infradistribution that is above  and , we must have equality, and

(and same for ) And we've shown the three definitions of the supremum are identical.

Proposition 16:


To recap, 

Now,  can be turned into a concave monotone functional , by LF-duality. Further, it's convex, closed, and upper-complete due to being the intersection of two convex closed upper-complete sets. Let's use  to refer to its corresponding functional. Then:


And the same applies to , and this applies to all functions, so  (and same for ).

We know from Proposition 15 that the least concave monotone functional above  and  is , so  (and same for ) Call the corresponding set of  as . Thus, translating this information ordering back to sets,

And same for . Therefore.

Therefore, all the subsets must be actual equalities, and so in particular we have:

Then we can go:


By  being equivalent to the infradistribution set induced by , expanding our definition of the sup, and translating back. And we're done!

Proposition 17: For any property in the table at the start of this section,  will fulfill the property if both components fulfill the property.

The way to show this is to use the alternate characterizations of supremum as intersection of the infradistribution sets, and the alternate characterizations of the various properties in terms of properties of minimal points.

We will make an observation used in all further proofs of properties. In order for  to have  in it, there must be a minimal point of the form  with  below it. Similarly, for  to contain , there must be a minimal point of the form  below it, with 

Thus, for  to lie in  and  and . Part of this is because said point lies in  and , the other part is because  is the lowest possible point in  associated with a measure component of , and it's the minimal. This observation will be used for all future sub-proofs in this proposition.

Homogenity: This is equivalent to "all minimal points have ", so if , then  (homogenity for ), and same for , so .

1-Lipschitzness: This is equivalent to "all minimal points have ", so if , then , and  (1-Lipschitzness of ), so .

Cohomogenity: This is equivalent to "all minimal points have ", so if , then , and  (cohomogenity of ), and , and , so . Then, .

C-additivity: This is equivalent to "all minimal points have ", so if , then , and  (C-additivity of ), so .

Crispness: This is equivalent to the conjunction of homogenity and C-additivity, both of which are preserved, so crispness is preserved as well.

Sharpness: Because all sharp infradistributions are crisp,  must be composed entirely of probability distributions if  and  are sharp. If any of the probability distributions in  aren't supported on  (the compact set associated with the sharp infradistribution ), then they aren't in , which is impossible. Symmetric arguments apply to . Thus,  only has probability distributions supported on . If there was any probability distribution supported on that set that was missing from , then it'd be present in  and , and thus present in , and minimal, so we have a contradiction. Therefore  consists of all probability distributions supported on  which is a compact set, so the supremum is sharp as well.

Proposition 18: If a family of infradistributions  is directifiable, then  (defined as the functional corresponding to the set ) exists and is an infradistribution. Further, for all conditions listed in the table, if all the  fulfill them, then  fulfills the same property.

A family of infradistributions being directifiable is equivalent to "for any collection of finitely many infradistributions, the supremum exists". We also know that the supremum is exactly equivalent to set intersection. So, we'll show that directifiability (any collection of finitely many infradistributions has a supremum) implies that the intersection of all the infradistribution sets has the exact properties of a set-form infradistribution.

We have six properties to check. Nonemptiness, normalization (the existence of a point , existence of a point  with  and nonexistence of points with ), closure, convexity, upper-completion, and compact-projection (the measure components of the infradistribution are contained in a compact set of measures).

For closure, it's the intersection of closed sets, so it's closed. For convexity, it's the intersection of convex sets, so it's convex. For upper-completion, it's the intersection of upper-complete sets, so it's upper-complete. For compact-projection, the measure components of the countable intersection are contained within the countable intersection of the sets of measure components, which is contained in a compact set, so it fulfills that property too.

This just leaves nonemptiness and normalization. We'll show normalization, which automatically implies nonemptiness. The nonexistence of points with  is definitely not preserved under intersection.

However, the compact-projection property means that for any infradistribution set , the intersection of it with the surface of a-measures where  is compact, so we're intersecting a bunch of compact sets. Due to the existence of supremum infradistributions for each collection of finitely many infradistributions (directifiability), we have the nonempty finite intersection property needed to conclude that the intersection of compact sets is nonempty. The same argument applies to the existence of a point with . The presence of those two points witnesses nonemptiness and normalization.

These are the last two conditions we needed to conclude the set represents an infradistribution, so the infinite supremum exists and is the infradistribution we need.

For preservation of the various properties, we can just reuse the arguments from Proposition 17 with only trivial modifications.

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