When does $\lim_{n\to\infty}f(x+\frac{1}{n})=f(x)$ a.e. fail We know that if $f\in L^1(\mathbb{R})$, then $\|f(\cdot+1/n)-f(\cdot)\|_{L^1}\to 0$ as $n\to \infty$, which implies that there exists a subsequence $f_{n_k}=f(x+\frac{1}{n_k})$ such that $f_{n_k}\to f$ a.e. 
My problem is that 
Is there any $f\in L^1(\mathbb{R})$  function such that the almost convergence result 
$$
f(x)=\lim_{n\to\infty}f(x+1/n)\quad \text{a.e.}
$$ 
is NOT valid?   I intend to believe that there exits such function. Otherwise, there would be a beautiful statement such that $f(x)=\lim_{n\to\infty}f(x+1/n)$ a.e. for any $f\in L^1(\mathbb{R})$. 
Thank you for the long commenting below @David C. Ullrich. This question has been reduce to find a compact set $K$ such that for any $x\in K$, there are infinite $n$ such that $x+\frac{1}{n}\not\in K$. 

The solution I have is a bit lengthy, so I will define f and leave out details. For $0<x<1$ $a_n(x)=[(n!)x]-n[(n-1)!x]$ ,[.] being the greatest integer function.The $x=\sum \frac {a_n(x)} {n!}$ and this representation is unique. Let $\epsilon >0$, $\frac 1 {2^{N}} <\epsilon$ and  $B_k=\{x:a_{n_k} =0\}$ where $n_k =2^{N+k}$. Let $f=I_A$ where A is the union of the sets $B_k$, $k \geq 1$. Then $f(x-1/n)$ fails to converge to f(x) on $A^{c}$ whose measure is at least $1-\epsilon$.  
