Let $p \in [1, \infty)$ and $Y:= \mathbb R^d$. Let $L^p_{\text{loc}} (Y)$ be the space of measurable functions $f:Y \to \mathbb R$ such that $$ \|f\|_{L^p_{\text{loc}}} := \sup_{y \in Y} \|1_{B(y, 1)} f\|_{L^p} < \infty, $$ where $B(y, 1)$ is the open unit ball centered at $y$. Then $(L^p_{\text{loc}} (Y), \|\cdot\|_{L^p_{\text{loc}}})$ is complete but not separable. Let $(y_m)$ be a countable dense subset of $Y$. The sphere has Lebesgue measure $0$. Then by dominated convergence theorem (DCT), $$ \|f\|_{L^p_{\text{loc}}} = \sup_{m \in \mathbb N} \|1_{B(y_m, 1)} f\|_{L^p} \quad \forall f \in L^p_{\text{loc}} (Y). $$
I would like to prove an analogue of the main result in this thread, i.e.,
Theorem Let $f_n, f \in L^p_{\text{loc}} (Y)$ such that $\|f_n-f\|_{L^p_{\text{loc}}} \to 0$. Then there is a subsequence $(n_k)$ such that $f_{n_k} \xrightarrow{k \to \infty} f$ a.e.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Let $f_{n,m} := f_n 1_{B(y_m, 1)}$ and $g_{m} := f 1_{B(y_m, 1)}$. For each $m \ge 1$, we have $\|f_{n, m}-g_m\|_{L^p} \xrightarrow{n \to \infty} 0$.
- There is a subsequence $\varphi_1$ of $\mathbb N$ such that $f_{\varphi_1 (n), 1} \xrightarrow{n \to \infty} g_1$ a.e. Clearly, $\|f_{\varphi_1 (n), 2} - g_2\|_{L^p} \xrightarrow{n \to \infty} 0$.
- There is a subsequence $\varphi_2$ of $\varphi_1$ such that $f_{\varphi_2 (n), 2} \xrightarrow{n \to \infty} g_2$ a.e. Clearly, $\|f_{\varphi_2 (n), 3} - g_3\|_{L^p} \xrightarrow{n \to \infty} 0$.
- ...
Recursively, there is a sequence $(\varphi_m)$ such that for all $m \ge 1$:
- $\varphi_1$ is a subsequence of $\mathbb N$,
- $\varphi_{m+1}$ is a subsequence of $\varphi_m$, and
- $f_{\varphi_m (n), m} \xrightarrow{n \to \infty} g_m$ a.e.
We define a subsequence $\psi$ of $\mathbb N$ by $\psi (m) := \varphi_m (m)$. Then $\psi$ is a subsequence of $\varphi_m$ for every $m \ge 1$. Clearly, $f_{\psi (n)} \xrightarrow{n \to \infty} f$ a.e. This completes the proof.
