Construction versus existence?

Shadow · **Posted:** Tue, 4 Oct 2011 19:55:59 UTC

I'm given a function $f:S\to\mathbb{R}$ and I'm asked to show every function $g:S\to\mathbb{R}$ is measurable relative to $(\sigma(f),\mathcal{B}(\mathbb{R}))$ iff $\exists h:\mathbb{R}\to\mathbb{R}$ which is measurable relative to $\mathcal{B}(\mathbb{R})$ and g=f^*(h)

. 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 $\forall A\in\mathcal{B}(\mathbb{R})\;\exists A'\in\mathcal{B}(\mathbb{R})$ such that $g^{-1}(A)=f^{-1}(A')$ .

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 $\mathcal{A}$ be the family of all

such that the condition holds and somehow show it must be everything by appealing to some property of $\sigma(f)$ or f^*

, but I cannot see how to quite work this one.

Partial construction (?):

Consider the partition of

formed by $$\coprod_{x\in\mathbb{R}} g^{-1}(\{x\})$ where I throw out empty sets as irrelevant. Then $\forall x\exists A_x\in\mathcal{B}(\mathbb{R})$ such that $g^{-1}(\{x\})=f^{-1}(A_x)$ , and because fibers commute with unions, we have: $$g^{-1}(\{x\})=\bigcup_{y\in A_x}f^{-1}(\{y\})$ , then I somehow get a partition of

by doing this for all points $x\in\mathbb{R}$ and disjointifying the A_x

by taking sufficient complements (which is fine, since we have a $\sigma$ -algebra) and somehow producing

from this by demanding $h(g^{-1}(\{x\}))\in A_x$ or even just $h(g^{-1}(\{x\}))=y$ for some choice of $y\in A_x$ which must be nonempty if $g^{-1}(\{x\})$ 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.

outermeasure · **Posted:** Tue, 4 Oct 2011 23:39:52 UTC

Shadow wrote:

I'm given a function $f:S\to\mathbb{R}$ and I'm asked to show every function $g:S\to\mathbb{R}$ is measurable relative to $(\sigma(f),\mathcal{B}(\mathbb{R}))$ iff $\exists h:\mathbb{R}\to\mathbb{R}$ which is measurable relative to $\mathcal{B}(\mathbb{R})$ and g=f^*(h)

. 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 $\forall A\in\mathcal{B}(\mathbb{R})\;\exists A'\in\mathcal{B}(\mathbb{R})$ such that $g^{-1}(A)=f^{-1}(A')$ .

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 $\mathcal{A}$ be the family of all

such that the condition holds and somehow show it must be everything by appealing to some property of $\sigma(f)$ or f^*

, but I cannot see how to quite work this one.

Partial construction (?):

Consider the partition of

formed by $$\coprod_{x\in\mathbb{R}} g^{-1}(\{x\})$ where I throw out empty sets as irrelevant. Then $\forall x\exists A_x\in\mathcal{B}(\mathbb{R})$ such that $g^{-1}(\{x\})=f^{-1}(A_x)$ , and because fibers commute with unions, we have: $$g^{-1}(\{x\})=\bigcup_{y\in A_x}f^{-1}(\{y\})$ , then I somehow get a partition of

by doing this for all points $x\in\mathbb{R}$ and disjointifying the A_x

by taking sufficient complements (which is fine, since we have a $\sigma$ -algebra) and somehow producing

from this by demanding $h(g^{-1}(\{x\}))\in A_x$ or even just $h(g^{-1}(\{x\}))=y$ for some choice of $y\in A_x$ which must be nonempty if $g^{-1}(\{x\})$ 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 $g^{-1}$ all that much. You can just construct h via a limit argument using the standard dyadic trick --- construct approximations h_n

of h corresponding to approximations $g_n=2^{-n}\lfloor 2^ng\rfloor$ 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:

Trap 2:

Spoiler:

The procedure in detail:

Spoiler:

Edit: typo corrected.

Shadow · **Posted:** Wed, 5 Oct 2011 23:08:59 UTC

outermeasure wrote:

Shadow wrote:

I'm given a function $f:S\to\mathbb{R}$ and I'm asked to show every function $g:S\to\mathbb{R}$ is measurable relative to $(\sigma(f),\mathcal{B}(\mathbb{R}))$ iff $\exists h:\mathbb{R}\to\mathbb{R}$ which is measurable relative to $\mathcal{B}(\mathbb{R})$ and g=f^*(h)

. 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 $\forall A\in\mathcal{B}(\mathbb{R})\;\exists A'\in\mathcal{B}(\mathbb{R})$ such that $g^{-1}(A)=f^{-1}(A')$ .

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 $\mathcal{A}$ be the family of all

such that the condition holds and somehow show it must be everything by appealing to some property of $\sigma(f)$ or f^*

, but I cannot see how to quite work this one.

Partial construction (?):

Consider the partition of

formed by $$\coprod_{x\in\mathbb{R}} g^{-1}(\{x\})$ where I throw out empty sets as irrelevant. Then $\forall x\exists A_x\in\mathcal{B}(\mathbb{R})$ such that $g^{-1}(\{x\})=f^{-1}(A_x)$ , and because fibers commute with unions, we have: $$g^{-1}(\{x\})=\bigcup_{y\in A_x}f^{-1}(\{y\})$ , then I somehow get a partition of

by doing this for all points $x\in\mathbb{R}$ and disjointifying the A_x

by taking sufficient complements (which is fine, since we have a $\sigma$ -algebra) and somehow producing

from this by demanding $h(g^{-1}(\{x\}))\in A_x$ or even just $h(g^{-1}(\{x\}))=y$ for some choice of $y\in A_x$ which must be nonempty if $g^{-1}(\{x\})$ 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 $g^{-1}$ all that much. You can just construct h via a limit argument using the standard dyadic trick --- construct approximations h_n

of h corresponding to approximations $g_n=2^{-n}\lfloor 2^ng\rfloor$ 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:

Trap 2:

Spoiler:

The procedure in detail:

Spoiler:

OK, so I think I mostly understand this, but in your notation: why should the $B_{j,n}$ not be disjoint? It looks like it will just be sets like $[{k\over 2^n},{k+1\over 2^n})$ which ought to be disjoint. And why $f(S)\subseteq B$ 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 $$h(x)=\lim_{n\to\infty} \chi_{_B}h_n(x)$ .

Thanks again.

outermeasure · **Posted:** Wed, 5 Oct 2011 23:14:51 UTC

Shadow wrote:

OK, so I think I mostly understand this, but in your notation: why should the $B_{j,n}$ not be disjoint? It looks like it will just be sets like $[{k\over 2^n},{k+1\over 2^n})$ which ought to be disjoint. And why $f(S)\subseteq B$ 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 $$h(x)=\lim_{n\to\infty} \chi_{_B}h_n(x)$ .

Thanks again.

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

Shadow · **Posted:** Wed, 5 Oct 2011 23:23:36 UTC

And how do we know the disjointification still gives us full image $A_{j,n}$ ? This one I feel must be obvious, but I'm missing it.

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

outermeasure · **Posted:** Wed, 5 Oct 2011 23:31:58 UTC

Shadow wrote:

And how do we know the disjointification still gives us full image $A_{j,n}$ ? This one I feel must be obvious, but I'm missing it.

Recall inverse image behaves under unions, complements, and intersections, so
$\begin{aligned} f^{-1}(B_{j,n}\cap(\bigcup_{i\neq j} B_{i,n})^c)&=f^{-1}(B_{j,n})\cap(\bigcup_{i\neq j} f^{-1}B_{i,n})^c\\ &=A_{j,n}\cap(\bigcup_{i\neq j} A_{i,n})^c\\ &=A_{j,n} \end{aligned}$ .

The condition $f(S)\subseteq B$ means at every point of f(S)

the

has a finite limit, which follows from $\lim f^*h_n(s)=\lim h_n(f(s))=\lim g_n(s)=g(s)$ for all $s\in S$ .

Shadow · **Posted:** Wed, 5 Oct 2011 23:32:46 UTC

outermeasure wrote:

Shadow wrote:

And how do we know the disjointification still gives us full image $A_{j,n}$ ? This one I feel must be obvious, but I'm missing it.

Recall inverse image behaves under unions, complements, and intersections, so $f^{-1}(B_{j,n}\cap(\bigcup B_{i,n})^c)=f^{-1}(B_{j,n})\cap(\bigcup f^{-1}B_{i,n})^c=A_{j,n}\cap(\bigcup A_{i,n})^c=A_{j,n}$ .

The condition $f(S)\subseteq B$ means at every point of f(S)

the

has a finite limit, which follows from $\lim f^*h_n(s)=\lim h_n(f(s))=\lim g_n(s)=g(s)$ for all $s\in S$ .

Ah, just changed it too. Sorry about that. Thanks for the $f(S)\subseteq B$ though, that was still eluding me.

outermeasure · **Posted:** Wed, 5 Oct 2011 23:50:03 UTC

Essentially the same proof gives the following version:

Suppose $X=(X,\mathcal{X})$ and $Y=(Y,\mathcal{Y})$ 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 $\sigma$ -algebra, $f\colon X\to Y$ measurable which induces the sigma algebra $\mathcal{X}_0$ on Y. Then a Z-valued $(\mathcal{Y},\mathcal{B}(Z))$ -measurable function g is $(\mathcal{X}_0,\mathcal{B}(Z))$ -measurable if and only if there exists measurable function $h\colon(Y,\mathcal{B})\to(Z,\mathcal{B}(Z))$ such that g(x)=h(f(x))

for all $x\in X$ .

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Construction versus existence?

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