Let X,Y to be mappings from the sample space Ω to R and suppose Y is measurable with respect to σ(X), the smallest σ-field that makes X measurable.
Does it follow that there exists some Borel-measurable function f: R → R such that Y=f(X)?
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$\begingroup$ The question seems quite limited in scope/interest. Is there a particular application you have in mind? $\endgroup$– Yemon ChoiNov 3, 2009 at 8:28
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6$\begingroup$ I think it's interesting, actually. Y being measurable with respect to sigma(X) is an abstract, high-level set-theoretic concept, while Y being f(X) for a function f is visceral and down-to-earth. The fact that the two are connected is important for intuition: Y being measurable with respect to sigma(X) essentially means that Y doesn't contain any information that X doesn't already have, so roughly, you can compute Y given X (ignoring all recursion-theoretic issues). $\endgroup$– Darsh RanjanNov 3, 2009 at 16:47
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1$\begingroup$ The requirement that one r.v. is measurable with respect to the smallest sigma field that makes another r.v. measurable appears quite often in probability theory; for example, its part of the definition of conditional expectation and stopping time. I'm trying to get a feel for what it means. $\endgroup$– alexNov 4, 2009 at 22:34
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$\begingroup$ This is called the Doob-Dynkin lemma, or Doob's functional representation lemma. This is more generally true when $Y$ takes its values in a Polish space and $X$ takes its values in any measurable space. See lemma 1.13 in Kallenberg's book Foundations of modern probability. $\endgroup$– Stéphane LaurentAug 14, 2018 at 15:51
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3 Answers
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The answer is yes. The proof is quite standard. 1. If Y=1_A, where A is in \simga(X), then by the definition of \sigma(X) there exists a Borel set B such that A = X^{-1}(B) and therefore
Y(\omega) = 1_A(\omega) = 1_B(X(\omega)) = f(X(\omega)),
where we put f:= 1_B (of course f is now a Borel function).
2. If now Y is a simple r.v. i.e. it can be written in form Y = \sum_{i=1}^n c_i 1_{A_i}, where A_i are sets in \sigma(X), then using the previous point we can find Borel functions f_i such that 1_{A_i} = f_i(X) and obviously in this case f = \sum_{i=1}^n c_i f_i.
3. Finally, any r.v. Y measurable w.r. to \sigma(X) can be approximated by a sequence of simple r.v. Y_n measurable w.r. to \sigma(X) i.e. Y_n -> Y almost surely. By the previous point there exist f_n such that Y_n = f_n(X). Now we can define f(x) = \lim_n f_n(x) if the limit exists and put f(x)=0 otherwise. It is easy to check that f is a Borel function (basically it is a limit of Borel functions), and that Y = f(X).
answered Nov 3, 2009 at 12:07
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1$\begingroup$ Don't we need the sequence Y_n to converge to Y pointwise, not just almost surely, since we want Y = f(X) pointwise, not just almost surely? Of course, this isn't a problem for the proof, since such a sequence exists. Also, a way to avoid the distinction between when lim_n f_n exists and when it doesn't is to just take f = limsup_n f_n. $\endgroup$ Jan 11, 2010 at 22:22
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If R is the reals, yes (as already explained). But if R is just some other Borel space, then perhaps not.
answered Nov 3, 2009 at 12:38
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2$\begingroup$ Do you have any counterexamples? Or do you just have worries about the generality of the proof? It looks like the important thing we need is just the ability to do infinite summations in R. $\endgroup$ Nov 6, 2009 at 6:47
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$\begingroup$ One could replace the reals by any standard Borel space (Borel sets of a complete and separable metric space), since such spaces are Borel isomorphic to the reals. General Borel spaces can be much less well behaved (things like uncountable products of the unit interval). Btw: The theorem asked for is known as the Doob-Dynkin Lemma: en.wikipedia.org/wiki/Doob%E2%80%93Dynkin_lemma $\endgroup$ Jul 24, 2010 at 10:27
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It is trivially true but maybe worth noting that the converse is also true - if there exists such an f, then Y is σ(X)-measurable.
This and the question asked are theorem 20.1(ii) in Billingsley's Probability and Measure, 3rd edition.
