Measurability of piecewise defined function on a product space

Question

Let

• $(\Omega,\mathcal A)$ and $(X,\mathcal X)$ be measurable spaces
• $g_n:\Omega\times X\to\mathbb R$ be $\mathcal A\otimes\mathcal X$-measurable for $n\in\mathbb N$

Assume that for all $x\in X$, there is a $N_x\in\mathcal A$ such that $(g_n(\omega,x))_{n\in\mathbb N}$ is convergent for all $\omega\in\Omega\setminus N_x$. Now, let $$g(\omega,x):=\begin{cases}\displaystyle\lim_{n\to\infty}g_n(\omega,x)&&\text{, if }\omega\in\Omega\setminus N_x\\0&&\text{, otherwise}\end{cases}$$ for $(\omega,x)\in\Omega\times X$. How can we conclude that $g$ is $\mathcal A\otimes\mathcal X$-measurable?

My problem with this task is the dependence of $N_x$ on $x$. How do we need to argue?

Answer 1

This is not true the way it is stated. Take $X$ any uncountable space with the $\sigma$-algebra of countable/co-countable subsets, and let $g_n=1$ on $X\times X$.

The diagonal $\Delta_X=\left\{(x,x):x\in X\right\}$ is not measurable, but we could choose $N_x=\Omega\setminus \left\{x\right\}$, so $g$ is the characteristic of the diagonal, not measurable.

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If we instead take $\Omega\setminus N_x=\left\{\omega:(g_n(\omega,x))_n\text{ converges}\right\}$ (so we are not making a choice depending on $x$), then $N_x$ is simply the $x$-section of the complement of $N'=\left\{(\omega,x):(g_n(\omega,x))\text{ converges}\right\}$, which is measurable.

This becomes a particular case of the following fact, with $\mathcal{Y}=\Omega\times X$:

• If $\mathcal{Y}$ is a measurable space and $f_n$ is a sequence of real measurable functions on $\mathcal{Y}$, then $C=\left\{y:(f_n(y))_n\text{ converges}\right\}$ is measurable, and $\lim_n f_n$ is measurable on $C$.