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Let $(X, \mathcal{M}, \mu)$ be a measure space and denote the Lebesgue measurable subsets of $\mathbb{R}$ by $\mathcal{L}$. Assume $f : X \times \mathbb{R} \rightarrow \mathbb{R}$ has the following properties: (i) $f(\cdot, t)$ is $\mathcal{M}$-measurable for all $t$, (ii) the map $t \mapsto \int_A f(x, t) \, d\mu(x)$ is Lebesgue measurable, (iii) $|f(x, t)| \le C$ for all $x \in X$, $t \in \mathbb{R}$.

Clearly, in this case the iterated integral $\int_B \int_A f(x, t) \, d\mu(x) \, dt$ exists for all $A \in \mathcal{M}$ and $B \in \mathcal{L}$ (we could even weaken the assumptions but I am happy with these). Can we conclude that $f$ is measurable with respect to the product $\sigma$-algebra $\mathcal{M} \otimes \mathcal{L}$?

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    $\begingroup$ You already have the answer from @geetha290krm. Let me also point out that, the classic example of the failure, under CH, of Tonelli theorem for non-measurable functions (see en.wikipedia.org/wiki/…), also shows that $f$ needs not even be measurable w.r.t. the completion of the product $\sigma$-algebra under the product measure. $\endgroup$
    – David Gao
    Commented Jun 17 at 4:46
  • $\begingroup$ Thank you @David Gao. Do you think we could get measurability with respect to the completion if in (ii) the map is continuous? Also, we can assume $X = \mathbb{R}^n$. $\endgroup$ Commented Jun 17 at 5:31
  • $\begingroup$ No. The counterexample already has the map in (ii) being continuous - in fact, constantly $0$. $\endgroup$
    – David Gao
    Commented Jun 17 at 5:35
  • $\begingroup$ @David Gao Yes but $F \times E$ is measurable in the completion isn't it? It's a set of measure $0$. $\endgroup$ Commented Jun 17 at 5:37
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    $\begingroup$ I’m not referring to @geetha290krm’s answer. I’m referring for the example I mentioned, the indicator function of a well-ordering of $\mathbb{R}$, assuming CH. That indicator function is not measurable, even under the completion. And $X = \mathbb{R}$ in that example. (The CH assumption is unnecessary. You can use similar arguments as in mathoverflow.net/a/447952/504602 to remove the CH assumption. It’s just easier to state under CH.) $\endgroup$
    – David Gao
    Commented Jun 17 at 5:54

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Counter-example: Let $E$ be a non-Lebesgue measurable set and $F$ be a measurable set of measure $0$ w.r.t. $\mu$. Let $f(x,y)=1$ of $x \in F$ and $y\in E$ and $0$ otherwise.

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  • $\begingroup$ Thanks! I would be interested in a positive result as well. In a book I am reading it is claimed that $f$ is measurable if $f(\cdot, t_k)$ converges weak$^*$ to $f(\cdot, t)$ in $L^\infty(X)$ whenever $t_k \to t$. This corresponds to the continuity of the map in (ii) rather than mere measurability. How can we prove this? Assume $X = \mathbb{R}^n$. $\endgroup$ Commented Jun 17 at 5:35
  • $\begingroup$ In regard to my new question, I meant measurability with respect to the completion of the product $\sigma$-algebra. I can't edit my previous comment. $\endgroup$ Commented Jun 17 at 5:47
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Let me just write down the counterexample I mentioned in comments, without assuming CH.

The function is the indicator function of the set constructed in this answer. To summarize, let $\kappa$ be the least cardinal of a set of real numbers which is not null. Accordingly, pick $Y \subset \mathbb{R}$ s.t. $|Y| = \kappa$ and $Y$ is not null. Fix a well-ordering $\prec$ of $Y$ of order type $\kappa$. Then let $E = \{(x, y) \in Y^2: x \prec y\}$, regarded as a subset of $\mathbb{R}^2$. Let $f$ be the indicator function $1_E$. Note that $(X, \mathcal{M}, \mu) = (\mathbb{R}, \mathcal{L}, m)$ in this example, where $\mathcal{L}$ is the $\sigma$-algebra of Lebesgue measurable sets and $m$ is the Lebesgue measure.

Observe that, for any fixed $t$, if $t \notin Y$, then $f(\cdot, t)$ is the zero function. If $t \in Y$, then $f(\cdot, t)$ is the indicator function of the set $\{y \in Y: y \prec t\}$, which has cardinality strictly less than $\kappa$, whence a null set. Either way, $f(\cdot, t)$ is measurable and is zero a.e. So, condition (i) is satisfied, and condition (ii) is satisfied - in fact, $\int_A f(x, t) \, dx = 0$ for any measurable $A \subset \mathbb{R}$ and any $t \in \mathbb{R}$, so the function in (ii) is continuous. (iii) is obviously satisfied with $C = 1$.

Now, we observe that $f$ is not measurable, even w.r.t. the completion of $\mathcal{L} \otimes \mathcal{L}$ under the product measure $m \otimes m$. Assume to the contrary that it is measurable. Then by Tonelli theorem, the iterated integrals exist and are equal. We have already seen that $\int f(x, t) \, dx = 0$ for all $t$, whence we must have $\int f(x, t) \, dt = 0$ for a.e.-$x$. As $Y$ is not null, there must exists some $x_0 \in Y$ s.t. $\int f(x_0, t) \, dt = 0$. But $f(x_0, \cdot)$ is the indicator function of $\{y \in Y: x_0 \prec y\}$, whence the set must be null. But the set $\{y \in Y: y \preceq x_0\}$ has cardinality strictly smaller than $\kappa$, whence null as well, so $Y = \{y \in Y: x_0 \prec y\} \cup \{y \in Y: y \preceq x_0\}$ is null, contradicting the assumptions on $Y$. Hence, $f$ cannot be measurable, even after completing the product $\sigma$-algebra.

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