Let $E \subset \mathbb{R}^d$ be a measurable set, and $1 \leq p \leq \infty$. Suppose $(f_n), (g_n)$ are two sequences in $L^p(E)$, $f_n \to f$ in $L^p(E)$, and $g_n \to g$ in $L^p(E)$. Prove that $h_n = \max\{f_n, g_n\}$ also converges in $L^p(E)$.
Proof
Here, my goal is to prove that for $h$ defined as $h = \max\{f, g\}$ we have
$$\lim_{n\to\infty}\|h - h_n\|_{L^p(E)} = 0$$
So, break $E$ into the union of two disjoint sets \begin{align} E &= E_{f_n}\cup E_{g_n} \quad \text{where} \\ \\ E_{f_n} &= \{x \in E: f_n(x) >= g_n(x)\} \quad \text{and} \\ \\ E_{g_n} &= \{x \in E: f_n(x) < g_n(x)\} \end{align}
Then $E_{f_n}$ and $E_{g_n}$ must be measurable (don't know how to prove that) and we have $h_n = f_n$ on $E_{f_n}$ and $h_n = g_n$ on $E_{g_n}$. Therefore
\begin{align} \left(\int_E \left|h_n\right|^p\right)^{1/p} = \left(\int_{E_{f_n}} \left|f_n\right|^p\right)^{1/p} + \left(\int_{E_{g_n}} \left|g_n\right|^p \right)^{1/p} \end{align}
which happens to show $h_n$ is in $L^p(E)$ for all $n \in \mathbb{N}$, since $f_n, g_n$ are. If we take $\chi_{E_{f_n}}$ to be the characteristic function for $f_n$ (and similarly for $g_n$) we have
\begin{align} \left(\int_E \left|h_n - h\right|^p\right)^{1/p} &= \left(\int_{E} \left|f_n - f\right|^p \chi_{E_{f_n}} \right)^{1/p} + \left(\int_{E} \left|g_n - g\right|^p \chi_{E_{g_n}} \right)^{1/p} \\ \\ &\leq \left(\int_{E} \left|f_n - f\right|^p \right)^{1/p} + \left(\int_{E} \left|g_n - g\right|^p \right)^{1/p} \end{align}
where the inequality is due to the monotonicity of Lebesgue integrals. Here, by hypothesis, the right side goes to zero as $n \to +\infty$, so by the squeeze theorem, $h_n$ is convergent in $L^p(E)$.
Discussion
The introduction of the characteristic function was unnecessary given monotonicity holds in the sense of sets and subsets so I could have just done that. Besides that, I'm still not sure if this proof is valid. I don't really have a good handle on what proof would be required to show a measurable set can be guaranteed to be broken into disjoint measurable sets with respect to the operation of max on measurable functions.
Another point that is required here, I believe is to show that the limit of a sequence of measurable functions is measurable. But I mean, how much stuff do I really have to prove to support this type of argument? Context -- functional analysis.
There is a much simpler argument: $\max \{f_n,g_n\}=\frac {f_n+g_n+|f_n-g_n|} 2$ so it is enough to show that $|f_n-g_n| \to |f-g|$. This follows from the fact that $||f_n-g_n|-|f-g||\leq |f_n-f|+|g_n-g|$ and $a^{p}+b^{p} \leq 2^{p} (a+b)$ for $a,b \geq 0$.