$L^p_{\text{loc}}$ convergence implies almost everywhere convergence Suppose $\{f_n\}\subset L^p(B)$ for the unit ball $B\subset\mathbb{R}^n$ converges in $L^p_{\text{loc}}$, i.e. $\int_V\lvert f_n(x)-f(x)\rvert^p dx\rightarrow 0$ for all $V\subset\subset B$. Can we say that $f_{n_i}(x)\rightarrow f(x)$ almost everywhere in $B$ for some subsequence $\{f_{n_i}\}$?
My idea for a proof was to consider a sequence of balls $B_{r_k}$ with $r_k\rightarrow 1$ and denote by $\{f^k_n\}$ the subsequence that converges almost everywhere in $B_{r_k}$. Then I thought I could pick a diagonal subsequence that converges almost everywhere in $B$. However I'm not sure how to show that the diagonal sequence converges almost everywhere.
Maybe this is not true, in that case can we find a counter-example?

You need to be careful about how exactly your subsequences are constructed.
Take a sequence $r_k\nearrow 1$ as $k\to\infty$, and let $B_k := B(0,r_k)$.
Let us start with $B_1$. By the assumptions there is a subsequence $f_{\sigma(1,1)}, f_{\sigma(1,2)}, f_{\sigma(1,3)},\dotsc$ which converges pointwise a.e. in $B_1$. Let $Z_1\subset B_1$ be the measure zero set where the pointwise convergence fails, so $(f_{\sigma(1,n)})$ converges pointwise everywhere in $B_1\setminus Z_1$ as $n\to\infty$.
Now, for $B_2$, we let $(f_{\sigma(2,n)})$ be a further subsequence of $(f_{\sigma(1,n)})$ that converges pointwise a.e. in $B_2$. This is possible since $(f_{\sigma(1,n)})$ certainly converges in $L_{\rm loc}^p$ as well. Continuing in this manner, we let $(f_{\sigma(k,n)})$ be a subsequence of the $(f_n)$ with the properties that

*

*$(f_{\sigma(k,n)})$ is a subsequence of $(f_{\sigma(k-1,n)})$

*$(f_{\sigma(k,n)})$ converges pointwise a.e. in $B_k$, with exceptional measure zero set $Z_k\subset B_k$
Let's arrange everything into a grid:
\begin{equation*}
\begin{array}{cccc}
f_{\sigma(1,1)} & f_{\sigma(1,2)} & f_{\sigma(1,3)} &\cdots      \\
f_{\sigma(2,1)} &f_{\sigma(2,2)} & f_{\sigma(2,3)} & \cdots      \\
f_{\sigma(3,1)} &f_{\sigma(3,2)} & f_{\sigma(3,3)} & \cdots      \\
\vdots     & \vdots     & \vdots     & \ddots
\end{array}
\end{equation*}
It's then not hard to see that the diagonal sequence $g_n = f_{\sigma(n,n)}$ is such that $(g_n)_{n\geq k}$ is a subsequence of $(f_{\sigma(k,n)})_{n\geq 1}$.
Let $Z = \bigcup_i Z_i$, and note that $Z$ is measure zero as the countable union of measure zero sets. Let $x\in B\setminus Z$. Then $x\in B_k\setminus Z_k$ for some $k$. For this $k$, we see that $g_n(x)\to f(x)$ as $n\to\infty$ since, for $n\geq k$, $g_n(x) = f_{\sigma(n,n)}(x)$ is a subsequence of $(f_{\sigma(k,n)}(x))$, the latter of which has limit $f(x)$ by construction. Thus the $(g_n)$ converge pointwise everywhere in $B\setminus Z$, and is thus the desired subsequence of the original $(f_n)$.
