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A function $f: X \rightarrow \mathbb{R}$ is measurable if and only if its graph is measurable, $\{(x, t): t=f(x)\} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$.

This is easy when we consider the points under the graph, but not sure how to do the graph itself

XiaoMem24
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    If it is indeed easy when we consider points under the graph, then it suffices to make the argument that your set can be written as $$ S = {(x,t) : t \leq f(x)} \setminus {(x,t): t < f(x)}, $$ and the difference of measurable sets is measurable. – Ben Grossmann Nov 11 '20 at 02:05
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    It is also easy to prove that if $f: X \rightarrow \mathbb{R}$ is measurable, then ${(x, t): t=f(x)} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$. – Ramiro Nov 11 '20 at 02:08
  • It indeed seems that the "only if" direction is significantly trickier – Ben Grossmann Nov 11 '20 at 02:22
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    @BenGrossmann I SUSPECT that the "only if" part, in the general case, is false. The results I know in this direction impose additional conditions on $X$ (for instance, that $X$ be a complete separable metric space). See, for instance, https://www.ams.org/journals/proc/1974-044-01/S0002-9939-1974-0335728-X/S0002-9939-1974-0335728-X.pdf – Ramiro Nov 11 '20 at 02:34
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    @Xiao If we are given that the region under the graph is measurable, then we could apply the argument given here. Is this what you meant by "this is easy when we consider the points under the graph"? – Ben Grossmann Nov 11 '20 at 02:35
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    @XiaoMem24 So that is what you meant. In any case, as the other commenter mentions, there is reason to believe that the a function might fail to be measurable even though its graph is measurable. Do you have any reason to believe that the "only if" that you're looking for is a true statement? – Ben Grossmann Nov 11 '20 at 03:11
  • @XioMem24 What are two texts that say that the "only if" part holds in the general case? Please, could you give those references? – Ramiro Nov 11 '20 at 03:30
  • @XioMem24 In both texts you mentioned, it is a just proposed exercise. As I said, I SUSPECT that the "only if" part, in the general case, is false. The results I know in this direction impose additional conditions on $X$ (for instance, that $X$ be a complete separable metric space). See, for instance, https://www.ams.org/journals/proc/1974-044-01/S0002-9939-1974-0335728-X/S0002-9939-1974-0335728-X.pdf – Ramiro Nov 11 '20 at 03:41
  • @XiaoMen24 Exactly. It is advanced and it is a peer reviewed paper published by AMS (American Mathematics Society) in May 1974. The final result (proposition 6 of section 3) is the "only if" part under the assumption that $X$ is a complete separable metric space. I don't believe AMS would accept to publish the paper, if the general case could be proved just as an exercise of product measure and iterate integral. – Ramiro Nov 11 '20 at 04:01

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It is also easy to prove that if $f: X \rightarrow \mathbb{R}$ is measurable, then $\{(x, t): t=f(x)\} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$.

Since $f: X \rightarrow \mathbb{R}$ is measurable, we have that $g: X \times \mathbb{R} \rightarrow \mathbb{R} $ defined by $g(x,y)= y-f(x)$ is $\mathcal{F} \otimes \mathcal{B}(\mathbb{R})$-measurable.

So $g^{-1}(\{0\}) \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$. But $g^{-1}(\{0\}) = \{(x, t): t=f(x)\}$.

So $\{(x, t): t=f(x)\} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$.

Ramiro
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