I asked the following question here.

Let $f: [0, 1]^2 \to \mathbb{R}$ be such that for every $x \in [0, 1]$ the function $y \to f(x, y)$ is Lebesgue measurable on $[0, 1]$ and for every $y \in [0, 1]$ the function $x \to f(x, y)$ is continuous on $[0, 1]$.

Is $f$ measurable with respect to the completion of the product $\sigma$-algebra $\mathcal{A} \times \mathcal{A}$ on $[0, 1]^2$?

Here $\mathcal{A}$ is the Lebesgue $\sigma$-algebra on $[0, 1]$.

John Dawkins gave the following answer.

Yes. For $n\in\Bbb N$ define $$ f_n(x,y)=\cases{f(k/n,y),&$(k-1)/k\le x<k/n, k=1,2,\ldots,n$\cr f(1,y),&$x=1$.\cr} $$ The function $f_n$ is $\mathcal A\otimes\mathcal A$-measurable (even $\mathcal B\otimes\mathcal A$-measurable, where $\mathcal B$ denotes the Borel subsets of $[0,1]$). Because $f_n$ converges pointwise to $f$, the function $f$ is $\mathcal B\otimes\mathcal A$-measurable.

$($The notation $\mathcal{A} \times \mathcal{A}$ is ambiguous: it is literally the Cartesian product of $\mathcal{A}$ with itself; namely $\{(A_1, A_2) : A_i \in \mathcal{A}\}$. The $\sigma$-field of interest here is $\sigma\{A_1 \times A_2 : A_i \in \mathcal{A}\}$, for which I prefer the notation $\mathcal{A} \otimes \mathcal{A}$.$)$

Why does it follow from $f_n$ converging pointwise to $f$ that $f$ is $\mathcal{B} \otimes \mathcal{A}$-measurable?


If $\mathfrak F$ is a $\sigma$-algebra on $X$ and $f_n$ is a sequence of $\mathfrak F$-measurable functions which converge pointwise to $f$, then $f$ is $\mathfrak F$-measurable.

Since under this hypothesis $$f=\limsup_{n\to\infty} f_n=\inf_{m\in\Bbb N}\sup_{n\ge m} f_n$$

And $\inf\limits_k f_k=-\sup\limits_k -f_k$, it suffices to show that $\sup$ of a sequence of $\mathfrak F$-measurable functions is $\mathfrak F$-measurable.

In fact, if $f_k$ is said sequence, $$\left\{x\in X\,:\,\sup_{k\in N} f_k(x)\le \alpha\right\}=\bigcap_{k\in\Bbb N}\left\{x\in X\,:\,f_k(x)\le \alpha\right\}$$

which is countable intersection of $\mathfrak F$-measurable sets.


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