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I was looking up Egorov's Theorem on Wikipedia. https://en.wikipedia.org/wiki/Egorov%27s_theorem

One of the conditions is that the functions attain values in a separable metric space M, and in the "Discussion of assumptions" section following the theorem, the following is stated: "The separability of the metric space is needed to make sure that for M-valued, measurable functions f and g, the distance d(f(x), g(x)) is again a measurable real-valued function of x."

Can someone explain how to prove the boldfaced statement?

[EDIT]To be precise, I want to know how to prove "If the metric space M is separable, and the functions f and g are measurable functions attaining values in M, then the distance function d(f(x),g(x)) is measurable."

Thanks!

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    $\begingroup$ There are a couple of things you could be asking for, and it might help to make your question more explicit. You want proof that if the metric space is separable, then that distance function is measurable? You want an example where that distance function is not measurable (and the metric space is nonseparable)? Or something else? $\endgroup$ Jul 4, 2013 at 21:37
  • $\begingroup$ Thanks for the response! I want proof that if the metric space is separable, then that distance function is measurable. $\endgroup$
    – Jarris
    Jul 4, 2013 at 21:41
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    $\begingroup$ This question on MathOverflow gives you some indications. $\endgroup$
    – Martin
    Jul 4, 2013 at 22:08
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    $\begingroup$ The measurable space in question seems to be arbitrary. So even if $d$ is continous from the product $M \times M$ it is not necessarily $ \Sigma \otimes \Sigma$ measurable. I can't see why the separability would guarantee the measurability though $\endgroup$
    – Bunder
    Jul 4, 2013 at 22:19

4 Answers 4

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A more elegant and intuitive way (that I've come up with upon finishing the direct proof, so I'll leave it anyway...) to show it is as follows: if $M$ is a separable metric space, it is second-countable.

From that it follows that open sets are Borel in $M^2$ (with respect to the square of Borel algebra of $M$; this is because we don't need uncountable unions to form open sets).

In particular, continuous functions from $M^2$ to $\bf R$, such as $d$, are Borel with respect to product algebra. On the other hand, $x\mapsto (f(x),g(x))$ is obviously measurable with respect to the product Borel algebra, so the composition of this and $d$ is measurable.

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  • $\begingroup$ I'm confused. Aren't open sets always Borel by definition? $\endgroup$
    – FShrike
    Nov 13, 2021 at 12:53
  • $\begingroup$ Or did you mean it implies $\mathcal{B}(M\times M)=\mathcal{B}(M)\times\mathcal{B}(M)$ because every Borel set of $M\times M$ can be reached from countable unions and products of Borel sets of $M$? $\endgroup$
    – FShrike
    Nov 13, 2021 at 12:54
  • $\begingroup$ @FShrike Yes. I should have written "measurable". It's a pretty old post though, I'm not sure if I should edit and bump it for such a trivial change. $\endgroup$
    – tomasz
    Nov 14, 2021 at 0:18
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(Note: this is a direct, but rather haphazard proof, I've posted another answer with a more intuitive one, but I've already written down this one, so I'll leave it here.)

Denote $d(f(x),g(x))$ by $h(x)$, and enumerate the countable dense subset of $M$ with $s_n$. Note that $h(x)=r$ if and only if you have $$ (\forall n\forall m)\;d(f(x),s_n)<1/m\implies r-1/m<d(g(x),s_n)<r+1/m $$ To show measurability of $h$, it's enough to show that preimages of closed intervals are measurable, or to show that their complements are. But by looking at the previous formula, you can see that $h(x)\notin [a,b]$ if and only if $$ (\exists n\exists m)\;d(f(x),s_n)<1/m\land (a-1/m\geq d(g(x),s_n)\lor d(g(x),s_n)\geq b+1/m) $$ So $X\setminus h^{-1}[[a,b]]$ is a countable union of measurable sets, and we're done.

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  • $\begingroup$ Thank you so much. I appreciate both this approach and the other one. They are very helpful. $\endgroup$
    – Jarris
    Jul 4, 2013 at 23:05
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Suppose that $\{y_n:n\in \mathbb{N}\}$ is a countable dense subset of the separable metric space $(M,d).$ For each $n\in \mathbb{N},$ let $F_n,G_n\colon X\to \left[0,\infty\right)$ be the measurable functions \begin{align*} F_n(x) &= d(f(x),y_n)\\ G_n(x) &= d(g(x),y_n). \end{align*} Indeed these functions are measurable because they are the compositions of the continuous function $d(\cdot,y_n)$ with the measurable functions $f$ and $g$, respectively. By the triangle inequality, $H\leq F_n+G_n$ for every $n\in \mathbb{N},$ making $H(x)$ a lower bound on the set $\{F_n(x)+G_n(x):n\in \mathbb{N}\}$ for every $x\in X.$ To see that $H(x)$ is the greatest lower bound on this set, consider $\varepsilon>0,$ and let $N\in \mathbb{N}$ be such that $d(g(x),y_N)<\frac{\varepsilon}{2},$ and hence $2d(g(x),y_N)<\varepsilon.$ Then \begin{align*} F_N(x)+G_N(x) &= d(f(x),y_N)+d(g(x),y_N)\\ &\leq d(f(x),g(x))+2d(g(x),y_N)\\ &<H(x)+\varepsilon, \end{align*} so that $H(x)+\varepsilon$ fails to be a lower bound on $\{F_n(x)+G_n(x):n\in \mathbb{N}\}.$ Since $\varepsilon>0$ was arbitrary, $$ H(x)=\inf_{n\in \mathbb{N}}\left(F_n(x)+G_n(x)\right). $$ This is true of all $x\in X,$ hence $H$ is the measurable function $\inf_{n\in \mathbb{N}}\left(F_n+G_n\right),$ where we recall that the pointwise infimum of a countable family of measurable functions is measurable.

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Here is a counterexample to your statement without separability.

Consider $Y$ a set with cardinality strictly larger than continuum, and $d$ the discrete metric. Then it can be shown that the diagonal $\Delta \in \mathcal{B}(Y\times Y)- \mathcal{B}(Y)\otimes \mathcal{B}(Y)$; this is known as the Nedoma's paradox cf. https://www.drmaciver.com/2006/04/journal-of-obscure-results-1-nedomas-pathology/. Now put $X=Y\times Y$ and $f,g:Y\times Y\to Y$ projections onto the first and second coordinates respectively. By construction, $f,g$ are measurable, but the set $\Delta=\{(x,y)\in Y\times Y: d(f(x,y),g(x,y))<1\}$ is not measurable.

Also a related question: Counterexample to Egorov for functions valued in non-separable metric space

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