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I am doing a problem in Measure, Integration & Real Analysis by Sheldon Axler on page 105.

Suppose $(X, \mathcal{S})$ is a measurable space and $f: X \to \mathbb{R}$ is a function. Let $\operatorname{graph}(f) \subset X \times \mathbb{R}$ denote the graph of $f$:

$$\text{graph}(f) = \{(x, f(x)) : x \in X\}$$

Let $\mathcal{B}$ denote the $\sigma$-algebra of Borel subsets of $\mathbb{R}$. Prove that $\operatorname{graph}(f) \in \mathcal{S} \otimes \mathcal{B}$ if and only if $f$ is an $\mathcal{S}$-measurable function.

I've already solved the if statement. But I can't find a way to solve the only if statement. I saw other similar posts, but I didn't find the only if statement.

For the if one, consider $$ E_k = \bigcup_{j=1}^{2^k k - 1} f^{-1}\left(\left[\frac{j}{2^k}, \frac{j+1}{2^k}\right]\right) \times \left[\frac{j}{2^k}, \frac{j+1}{2^k}\right], \quad F_k = f^{-1}\left([2^k k, \infty]\right) \times [k 2^k, \infty), \quad G_k = E_k \cup F_k. $$ and we take the limit of $G_k$.

Now how about the only if statement? The difficult part is so show that $f^{-1}(B)\in \mathcal{S}$ for any Borel set. I'm struggling with this.

Update: My attempt: using $\text{graph}(f)\in \mathcal{S}\otimes\mathcal{B}$, construct an $\mathcal{S}\otimes\mathcal{B}$-measurable function. Then take its section, for example $g_y$, that could possibly be $f$, and we now that the this section is $\mathcal{S}$-measurable.

zzivv
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  • Have you checked this post? Or this one? – Daan May 21 '25 at 07:30
  • @Daan: I think that these post only show that the measurability of the function implies the measurability of the graph. – gerw May 21 '25 at 08:40
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    As far as I know, this direction is very hard, even in case that $\mathcal S = \mathcal B$. I do not know an easy proof. – gerw May 21 '25 at 08:40
  • Which version of the book you are using? I have checked the latest version on https://measure.axler.net/. The exercise appears as exercise 5B.4 on p. 135 and only requires you to show "if" and not "if and only if". The change to the exercise is also mentioned in the Erratum. – gerw May 21 '25 at 08:44
  • @gerw I might have used previous version. The author deleted the only if direction in the latest version so I guess it's not an easy one. – zzivv May 21 '25 at 09:25
  • It might be 2020's edition, the GTM one. – zzivv May 21 '25 at 09:31
  • Apart from being non-trivial, the only if part may require some hypothesis on $(X,\mathcal S)$ ($X$ Polish, $\mathcal S=$ Borel $\sigma-$ algebra). – Kavi Rama Murthy May 21 '25 at 10:04
  • There are examples of non measurable functions with measurable graphs. See https://mathoverflow.net/questions/376493/is-there-an-example-of-a-non-measurable-function-with-a-measurable-graph – Ramiro May 22 '25 at 03:38
  • Note that, assuming that $X$ is a Polish space and $\mathcal{S}$ is its Borel $\sigma$-algebra, if $\operatorname{graph}(f) \in \mathcal{S} \otimes \mathcal{B}(\mathbb{R})$ then $f$ is $\mathcal{S}$-$\mathcal{B}(\mathbb{R})$ measurable. The proof is more advanced and uses analytic sets. One such proof can found in section 3 proposition 6 of the paper : https://www.ams.org/journals/proc/1974-044-01/S0002-9939-1974-0335728-X/S0002-9939-1974-0335728-X.pdf – Ramiro May 22 '25 at 04:10

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