TITLE: When does $\lim_{n\to\infty}f(x+\frac{1}{n})=f(x)$ a.e. fail
QUESTION [21 upvotes]: 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 $x+\frac{1}{n}\not\in K$.

REPLY [4 votes]: Heh, people banging head against wall for days, then overnight three apparently correct answers appear. On the one hand, Giuseppe Negro's answer seems the simplest, but on the other hand it depends on Stein's theorem about maximal functions.
The point to this answer is to show that that other answer can easily be converted to an explicit counterexample, based on a vague recollection of how the proof of Stein's theorem goes.
Say $X_1,\dots$ are iid random variables, uniformly distributed on $[0,1]$. Define $$I_k=[X_k,X_k+1/k^2].$$Choose $N$ so $$\sum_N^\infty\frac1{k^2}<1,$$and let $$E=\bigcup_N^\infty I_k$$and $$f=\chi_E.$$
Let $$A_k=\bigcup_{\frac k2\le n\le k}\left(I_k-\frac1n\right).$$
As Giuseppe explains, $$m(A_k)\ge c/k.$$ So Borel-Cantelli shows that almost surely almost every point lies in infinitely many $A_k$. In particular, if $B$ is the set of $x\in[0,1]\setminus E$ such that $x$ lies in infinitely many $A_k$ then $m(B)>0$. But if $x\in B$ then $f(x)\ne\lim f(x+1/n)$.
(In case it's not clear, the point to adding the restriction $n\ge k/2$ to GN's definition of $A_k$ is to ensure that if $x\in A_k$ for infinitely many $k$ then $x\in I_k-1/n$, where $n$ is unbounded.)
Note the appplication of Borel-Cantelli is a little iffy, because $A_k$ need not be a subset of $[0,1]$. If that bothers you, regard the above as a contruction on the torus $\Bbb T=\Bbb R/\Bbb Z$. As Guiseppe points out, a counterexample on $\Bbb T$ implies a counterexample on $\Bbb R$.
Edit: It's been objected that the events $A_k$ are not independent. I didn't intend to give the impression I was saying that they were. In fact $A_k\subset\Bbb T$; if we regard $\Bbb T$ as a probability space then the $A_k$ are in fact "events", but regardless they're not the independent events that I was applying Borel-Cantelli to.
With  apologies for not explaining the following at the start -  I thought it was well known that if $\sum m(A_k)=\infty$ and you consider "random" translates then you almost surely cover almost every point infinitely often:
For each $k$ and each $x\in\Bbb T$ let $E_{k,x}$ be the event $$x\in A_k.$$That's an honest event, a subset of the probability space where the $X_k$ live. Now if $x$ is fixed it's clear that the events $E_{k,x}$ are independent (since, again for fixed $x$, there exists a set $S_k$ such that $x\in A_k$ if and only if $X_k\in S_k$). And since $X_k$ is uniformly distributed on $\Bbb T$ it's clear that $P(E_{x,k})=m(A_k)$. So for each $x$ it is almost surely true that $x$ lies in infinitely many $A_k$, by Borel-Cantelli. Now Fubini shows that it is almost surely true that almost every $x$ lies in infinitely many $A_k$.