Does the existence of the iterated integral imply measurability?

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 $$\mathcal{M} \otimes \mathcal{L}$$?

• 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. Commented Jun 17 at 4:46
• 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 $X = \mathbb{R}^n$. Commented Jun 17 at 5:31
• No. The counterexample already has the map in (ii) being continuous - in fact, constantly $0$. Commented Jun 17 at 5:35
• @David Gao Yes but $F \times E$ is measurable in the completion isn't it? It's a set of measure $0$. Commented Jun 17 at 5:37
• 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.) Commented Jun 17 at 5:54

2 Answers

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.

• 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 $X = \mathbb{R}^n$. Commented Jun 17 at 5:35
• 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. Commented Jun 17 at 5:47

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.