For $E$ a measurable set and $1\leq p<\infty $, assume $f_n\to f$ in $L^p(E)$. Show that there is a subsequence $\{f_{n_k}\}$ and a function $g\in L^p(E)$ such that $|f_{n_k}|\leq g$ almost everywhere on $E$ for all $k$.
By convergence in $L^p(E)$, I mean that $\|f_n-f\|_p\to 0$. I know that a convergent sequence in $L^p$ has a subsequence that converges pointwise a.e. to $f$ on $E$. Since $f\in L^p(E)$, it must be finite almost everywhere. I would like to define $g$ to be the pointwise supremum of the $f_{n_k}$, but there is no guarantee that this supremum is finite almost everywhere. How can I construct $g$ to bound this subsequence above pointwise?
Since $\|f_n-f\|_p\to 0$, for every $k\in \Bbb N$, there exists $n_k\in \Bbb N$, such that $\|f_{n_k}-f\|_p\le \frac{1}{2^k}$. Let $$g:=|f|+\sum_{k=1}^\infty |f_{n_k}-f|.$$ By definition, $g$ is measurable, and $g\ge |f|+|f_{n_k}-f|\ge|f_{n_k}|$ for every $k\in \Bbb N$ . Moreover, by Minkowski's inequality, $$\|g\|_p\le \|f\|_p+\sum_{k=1}^\infty \|f_{n_k}-f\|_p\le \|f\|_p+1.$$