Does a monotone convergence theorem for $L^\infty$ norm holds? Let $0 \leq f_n$ be a non-decreasing sequence on measurable functions converging to $f$. Does $|f_n|_{L^\infty} $ converges to $|f|_{L^\infty}$?
Motivation for the question:
I wanted to prove that if $f:X \times Y \to \mathbb{C}$ is measurable, then $|f(\cdot, y)|_{L^\infty}$ is measurable on $Y$ (when $X$ and $Y$ are sigma-finite). It is easy to prove this for simple functions, so I wanted to use MCT to conclude the proof.
 A: YES.
First of all, since $f_n$ is non-negative and non-decreasing, so is $|f_n|$.
Hence
$$
|f_1|_\infty\le |f_2|_\infty\le \cdots\le |f_n|_\infty \le \sup_{n\in\mathbb N}|f_n|_\infty=\lim_{n\to\infty}|f_n|_\infty
$$
Meanwhile
$$
f_n(x)\le f(x),\quad\text{for all $n\in\mathbb N$ and $n\in\mathbb N$}, 
$$
and hence $|f_n|_\infty\le|f|_\infty$, and consequently 
$$
\lim_{n\to\infty}|f_n|_\infty\le |f|_\infty\tag{1}
$$
If $\lim_{n\to\infty}|f_n|_\infty=\infty$, then clearly 
$\lim_{n\to\infty}|f_n|_\infty= |f|_\infty$. 
Assume now that $\lim_{n\to\infty}|f_n|_\infty=M<\infty$.
As $|f_n|_\infty$ is non-decreasing, this means that for every $n\in\mathbb N$, 
$$
0\le f_n(x)\le M, \quad\text{for all $x\in X$}. 
$$
Hence
$$
0\le f(x)=\lim_{n\to\infty}f_n(x)\le M, \quad\text{for all $x\in X$}. 
$$
and finally 
$$
|f|_\infty\le M=\lim_{n\to\infty}|f_n|_\infty. \tag{2}
$$
Now combine $(1)$ and $(2)$.
A: The answer is YES. Clearly, $f_n$ increasing  to $f$ implies $\|f_n\|_{\infty} \leq \|f\|_{\infty}$. To prove the other way note that $|f(x)| \leq \sup |f_n(x)| \leq \sup \|f_n\|_{\infty}$ almost everywhere. Hence $\|f\|_{\infty} \leq \sup \|f_n\|_{\infty}=\lim \|f_n\|_{\infty}$.
