Problem: $f$ is integrable on $\mathbb{R}$, prove there exists $h_n\to0$ where $h_n\in(0,1)$ such that $f(x+h_n)\to f(x)\ a.e\ x\in\mathbb{R}$
I tried to prove $\sum_{n=1}^\infty\mid f(x+h_n)-f(x)\mid<\infty\ a.e\ x\in\mathbb{R}$, but I got stuck. I'm not sure if it's a promising way.
Suggestion:
In a previous question, it was establesihed that for for any set of finite measure $E$, $\lim_{h\rightarrow0}\mu((E+h)\Delta E)=\|\mathbb{1}_{E+h}-\mathbb{1}_E\|_1\rightarrow0$ as $h\rightarrow0$.
Using (1), one can show that for any integrable simple function $s$, $\|s(\cdot -h)-s\|_1=\int|s(x-h)-s(x)|\,dx\xrightarrow{h\rightarrow0}0$
Any function $f\in L_1$ can be approximate by integrable simple function. That is, for any $\varepsilon>0$, there is an integrable simple function $s$ such that $\|f-s\|_1<\varepsilon$
Using (2) and (3) along with \begin{aligned} \|f(\cdot-h)-f\|_1&\leq\|f(\cdot-h)-s(\cdot+h)\|_1+\|s(\cdot-h)-s\|_1\\ &=\|f-s\|_1+\|s(\cdot-h)-s\|_1 \end{aligned} one can show that $\lim_{h\rightarrow}\|f(\cdot-h)-f\|_1=0$. (Here the translation invariance property of Lebesgue;s measure plays a role)
Back to the original problem. suppose $h_n\rightarrow0$ in $\mathbb{R}$. By (4) $\lim_n\|f(\cdot-h_n)-f\|_1=0$. By a well known result, one can extract a subsequence $h_{n_k}$ along which $f(\cdot+h_{n_k})-f(\cdot)$ converges almost surely.