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My question is : is there a $L^1(\mathbf R)$ function $f$ and a sequence $(t_n)_n$ of reals with limit $0$ such that $f_{t_n}=f(\cdot-t_n)$ does not converges to $f$ a.e. ?

There are several variations of this questions which are pretty much equivalent : one can drop the $L^1$ hypothesis and just assume mesurability of $f$, one could also assume that $f$ is the indicator function of a measurable subset of $[0;1]$.

What i already know :

The first thing to say is that $f_\tau$ converges to $f$ in $L^1$ norm when $\tau$ goes to zero. One way to prove that is to say that this is true for compactly suported continuous functions (because of absolute continuity) and conclude by density.

Since the $f_{t_n}$ are converging in $L^1$ norm there exists a subsequence $t_{\sigma(n)}$ such that $f_{t_{\sigma(n)}}$ converges almost everywhere to $f$. However one cannot directly use this result to conclude with the convergence a.e. of $f_{t_n}$, indeed there are many examples of sequences of functions that are converging in $L^1$ norm but not convergent a.e. In fact this show that the a.e. convergence does not corespond to any topological convergence, since the caracterisation of convergence with sub-sub-sequences doesn't apply here.

Another thing to note is that if $f$ is continuous at $x$ then $f_{t_n}(x)\to f(x)$ automatically, so if we search some $f$ such that $f_{t_n}$ does not converges a.e. to $f$ it has to be a function discontinuous on a set with positive measure. In fact for every function $g=0$ a.e. one must have that $f+g$ is discontinuous on a set with positive measure.

I believe the answer to my question is yes, but it's mostly an intuition, i don't even have euristic arguments for that. So i tried to find an example, the two kind of $f$ i have thought of are :

$f=\chi_{C}$ where $C$ is a fat cantor set or something like this kind of set and

$$f(x)=\chi_{[0;1]}\sum_{k=0}^\infty \frac{2^{-k}}{|x-q_k|^{\alpha}}$$ where $q_n$ is an enumeration of the rationals of $[0;1]$.

The problem is that it's hard to find the behavior of $|f_{t_n}(x)-f(x)|$ for general $x$ and such patological functions...

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  • $\begingroup$ something does not look right between $L^1$ and $\mathcal L^1$ $\endgroup$ – marmouset Sep 9 '16 at 13:47
  • $\begingroup$ It's $L^1$, not $\mathcal L^1$, that's why i use sequence of $t_n$ and not just $t\to 0$. By using sequences instead of $t\to 0$ the choice of representative for $f$ doesn't change the convergence ae property. Note that the only time i don't use a sequence and use $\tau\to 0$ this is not a problem since i'm talking about norm convergence. $\endgroup$ – Renart Sep 9 '16 at 14:01
  • $\begingroup$ Correct, and beautiful. $\endgroup$ – marmouset Sep 9 '16 at 14:22
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Yes. Consider $f(x)=\sum 2^{-n}g(x-q_n)$, where $q_n$ is an enumeration of $\mathbb Q$, and $g\in L^1$ is an unbounded non-negative bump function such as $g(x)=\chi_{(-1,1)}(x)|x|^{-1/2}$.

I'm now going to produce a sequence $t_n\to 0$ for which $f(x-t_n)$ is unbounded for every $x\in [-1,1]$. My first bunch of $t_n$'s will satisfy $|t_n|\le 1$. Start out by locating a $q_n$ close to $-1$ ($q_n=-1$ would work nicely). The corresponding bump $2^{-n}g(x-q_n)$ will be $\ge 1$, say, on an interval centered at $q_n$. By shifting this bump around, I can make sure that $f(x-t_n)\ge 1$ for every $x\in [-1,0]$ for some $n$. Cover the remainder of $[-1,1]$ in the same way.

Let's summarize: I have chosen finitely many elements $t_1, \ldots , t_N$ of my sequence, in such a way that $f(x-t_n)\ge 1$ for every $x\in [-1,1]$ for some $1\le n\le N$. Moreover, these satisfy $|t_n|\le 1$.

Now just continue in this way: In the second step, find the next batch of $t$'s, such that $f(x-t_n)\ge 2$ for all $x\in [-1,1]$ for some $n$, and $|t_n|\le 1/2$ for these $t_n$ etc.

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  • $\begingroup$ Ah you're right ! It's very close to the "wandering rectangle" sequences that converges in probability but not almost surely in fact. I'm surprised that your solution mimic this wandering rectangle patern, i didn't thought this was possible. Nice answer ! $\endgroup$ – Renart Sep 10 '16 at 17:10

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