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 Post subject: Construction versus existence?Posted: Tue, 4 Oct 2011 19:55:59 UTC
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I'm given a function and I'm asked to show every function is measurable relative to iff which is measurable relative to and . One direction is trivial since the composition of measurable functions is measurable. Intuitively it makes sense as well, but I'm a little stuck on the other.

I know by definition measurable in the sense indicated means such that .

And I feel it must only be a hop, skip, and a jump away from actually defining an with this relation since point fibers partition off the domain, , however I cannot quite see how to do it, so I think perhaps there is something sneakier I can do, such as letting be the family of all such that the condition holds and somehow show it must be everything by appealing to some property of or , but I cannot see how to quite work this one.

Partial construction (?):

Consider the partition of formed by where I throw out empty sets as irrelevant. Then such that , and because fibers commute with unions, we have: , then I somehow get a partition of by doing this for all points and disjointifying the by taking sufficient complements (which is fine, since we have a -algebra) and somehow producing from this by demanding or even just for some choice of which must be nonempty if is. The axiom of choice should ensure that this works. Does this function as a workable argument? It feels a bit ad-hoc, but who knows. Any help appreciated.

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 Post subject: Re: Construction versus existence?Posted: Tue, 4 Oct 2011 23:39:52 UTC
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I'm given a function and I'm asked to show every function is measurable relative to iff which is measurable relative to and . One direction is trivial since the composition of measurable functions is measurable. Intuitively it makes sense as well, but I'm a little stuck on the other.

I know by definition measurable in the sense indicated means such that .

And I feel it must only be a hop, skip, and a jump away from actually defining an with this relation since point fibers partition off the domain, , however I cannot quite see how to do it, so I think perhaps there is something sneakier I can do, such as letting be the family of all such that the condition holds and somehow show it must be everything by appealing to some property of or , but I cannot see how to quite work this one.

Partial construction (?):

Consider the partition of formed by where I throw out empty sets as irrelevant. Then such that , and because fibers commute with unions, we have: , then I somehow get a partition of by doing this for all points and disjointifying the by taking sufficient complements (which is fine, since we have a -algebra) and somehow producing from this by demanding or even just for some choice of which must be nonempty if is. The axiom of choice should ensure that this works. Does this function as a workable argument? It feels a bit ad-hoc, but who knows. Any help appreciated.

No AC is needed, and you don't actually want to consider all that much. You can just construct h via a limit argument using the standard dyadic trick --- construct approximations of h corresponding to approximations of g, and then take the limit of h_n. There are a couple of traps along the way for the unenlightened.

Trap 1:
Spoiler:
You need to make sure you choose pairwise disjoint sets before you try to define h_n on those sets.

Trap 2:
Spoiler:
You need to make sure behaves, and doesn't spoil .

The procedure in detail:
Spoiler:
The sets , for and fixed n, partitions into disjoint subsets using . Let such that . Now may not be pairwise disjoint, but you can replace them with which are pairwise disjoint Borels (remember for fixed n) with . Hence we have approximations

Obviously is Borel-measurable, so the set is also Borel, hence

is Borel. All that remains is to check that , which is easy.

Edit: typo corrected.

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 Post subject: Re: Construction versus existence?Posted: Wed, 5 Oct 2011 23:08:59 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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outermeasure wrote:
I'm given a function and I'm asked to show every function is measurable relative to iff which is measurable relative to and . One direction is trivial since the composition of measurable functions is measurable. Intuitively it makes sense as well, but I'm a little stuck on the other.

I know by definition measurable in the sense indicated means such that .

And I feel it must only be a hop, skip, and a jump away from actually defining an with this relation since point fibers partition off the domain, , however I cannot quite see how to do it, so I think perhaps there is something sneakier I can do, such as letting be the family of all such that the condition holds and somehow show it must be everything by appealing to some property of or , but I cannot see how to quite work this one.

Partial construction (?):

Consider the partition of formed by where I throw out empty sets as irrelevant. Then such that , and because fibers commute with unions, we have: , then I somehow get a partition of by doing this for all points and disjointifying the by taking sufficient complements (which is fine, since we have a -algebra) and somehow producing from this by demanding or even just for some choice of which must be nonempty if is. The axiom of choice should ensure that this works. Does this function as a workable argument? It feels a bit ad-hoc, but who knows. Any help appreciated.

No AC is needed, and you don't actually want to consider all that much. You can just construct h via a limit argument using the standard dyadic trick --- construct approximations of h corresponding to approximations of g, and then take the limit of h_n. There are a couple of traps along the way for the unenlightened.

Trap 1:
Spoiler:
You need to make sure you choose pairwise disjoint sets before you try to define h_n on those sets.

Trap 2:
Spoiler:
You need to make sure behaves, and doesn't spoil .

The procedure in detail:
Spoiler:
The sets , for and fixed n, partitions into disjoint subsets using . Let such that . Now may not be pairwise disjoint, but you can replace them with which are pairwise disjoint Borels (remember for fixed n) with . Hence we have approximations

Obviously is Borel-measurable, so the set is also Borel, hence

is Borel. All that remains is to check that , which is easy.

OK, so I think I mostly understand this, but in your notation: why should the not be disjoint? It looks like it will just be sets like which ought to be disjoint. And why is still somewhat eluding me. Also, I wrote this with characteristic functions instead of a piecewise definition, presumably that's all good and kosher? In particular I have .

Thanks again.

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 Post subject: Re: Construction versus existence?Posted: Wed, 5 Oct 2011 23:14:51 UTC
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OK, so I think I mostly understand this, but in your notation: why should the not be disjoint? It looks like it will just be sets like which ought to be disjoint. And why is still somewhat eluding me. Also, I wrote this with characteristic functions instead of a piecewise definition, presumably that's all good and kosher? In particular I have .

Thanks again.

Oops, some of the g should be f, corrected.

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 Post subject: Re: Construction versus existence?Posted: Wed, 5 Oct 2011 23:23:36 UTC
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And how do we know the disjointification still gives us full image ? This one I feel must be obvious, but I'm missing it.

EDIT: Ignore that, inverse images commute with intersections, I get it.

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 Post subject: Re: Construction versus existence?Posted: Wed, 5 Oct 2011 23:31:58 UTC
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And how do we know the disjointification still gives us full image ? This one I feel must be obvious, but I'm missing it.

Recall inverse image behaves under unions, complements, and intersections, so
.

The condition means at every point of the has a finite limit, which follows from for all .

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 Post subject: Re: Construction versus existence?Posted: Wed, 5 Oct 2011 23:32:46 UTC
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outermeasure wrote:
And how do we know the disjointification still gives us full image ? This one I feel must be obvious, but I'm missing it.

Recall inverse image behaves under unions, complements, and intersections, so .

The condition means at every point of the has a finite limit, which follows from for all .

Ah, just changed it too. Sorry about that. Thanks for the though, that was still eluding me.

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 Post subject: Re: Construction versus existence?Posted: Wed, 5 Oct 2011 23:50:03 UTC
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Essentially the same proof gives the following version:

Suppose and are measurable space, Z a complete separable metric space (actually a Polish space would do, we don't care too much about the metric) with the Borel -algebra, measurable which induces the sigma algebra on Y. Then a Z-valued -measurable function g is -measurable if and only if there exists measurable function such that for all .

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