Convergence in norm but not almost everywhere of $f_{t_n}=f(\cdot-t_n)$ to $f$ 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...
 A: Your idea of a fat Cantor set works too.  I'll repeat a construction I used
in this answer.
Let $C$ be a compact nowhere dense set of positive measure.  We can construct a sequence $t_n \to 0$ such that for every $x \in C$, there are infinitely many $n$ for which $x - t_n \notin C$.  In particular we have $\liminf_{n \to \infty} \chi_C(x-t_n) = 0$ and therefore $\chi_C(x-t_n) \not\to 1 = \chi_C(x)$.  Since $C$ has positive measure, we therefore do not have $\chi_C(\cdot-t_n) \to \chi_C$ a.e.
To construct the sequence, let $U = C^c$ and note that $U$ is open and dense.  So for each integer $m > 0$ and each $x \in C$, there exists $u \in U$ such that $|x-u| < 1/m$, which is to say that $x \in U+(x-u)$.  In other words, the collection of open sets $\{U+s : |s| < 1/m\}$ cover $C$ (indeed they cover $\mathbb{R}$).  Since $C$ is compact, there is a finite subcover $\{U+s_{m,1},\dots, U+s_{m, k_m}\}$.  That is to say, for every $x \in C$ there is some $1 \le i \le k_m$ for which $x - s_{m, i} \in U = C^c$.
Now let $(t_n)$ be the sequence
$$(s_{1,1}, \dots, s_{1, k_1}, s_{2,1}, \dots, s_{2, k_2}, \dots).$$
Since $|s_{i,m}| \le 1/m$ for every $m$, we have $t_n \to 0$, and there are infinitely many $t_n$ such that $x-t_n \notin C$ (namely, at least one of the $s_{m,i}$ for each $m$).
A: 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.
A: Here is a different solution to a slightly general problem that says that for any sequence $t_n\searrow 0$, and any $p\geq1$, there is a function $f\in L_p([0,1],m)$ ($m$ is Lebesgue measure) such that
$$m\big(\{x\in I:f(x-t_n)\stackrel{n}{\nrightarrow}f(x)\})>0$$

The solution is based on a nice result by Stein that considers linear operators on $L_p$ of the form
$$ T_nf=f*\mu_n$$
where $\mu_n$ is a sequence of finite measure supported on a compact set $K$ ($[0,1]$ for example).
It is easy to check that $\|T_nf\|_p\leq\mu_n(K)\|f\|_p$ for any $f\in L_p(\mathbb{R})$. Define the maximal function
$$ Mf(x):=\sup_n|T_nf(x)|$$
The result states the following:

Theorem (Stein). If $m\big(x:Mf(x)<\infty\big)>0 $ for all $f\in L_p(\mathbb{R},m)$, then there exists a constant $A>0$ such that
$$ \begin{align}m\big(\{x:Mf(x)>\lambda\}\big)\leq\frac{A}{\lambda^p}\|f\|^p_p\tag{1}\label{one}\end{align}$$
for all  $f\in L_p(\mathbb{R})$

Observation: We stated Stein's result as  real line result here for convenience, but as we will see in the explanation of the result, one can restrict to $L_p([0,1],m)$; moreover, the result is valid for $L(\mathbb{R}^d)$.

Solution to the OP:  Without loss of generality, assume that $0<t_n<1$. Then
$$ T_nf(x)=f(x-t_n)=f*\delta_{t_n}(x)$$
where $\delta_{x}$ is Dirac's measure concentrated at $x$, i.e. $\delta_x(A)=\mathbb{1}_A(x)$ for any Borel set $A$.
Notice that these are of the type of operators in Stein's result. If it were the case that for any $f\in L_1([0,1])$, $\lim_nf(x-t_n)=f(x)$ a.s., then $Mf(x)=\sup_n|f(x-t_n)|<\infty$ for almost surely all $x\in [0,1]$ and so, $Mf$ would satisfy an inequality of the form \eqref{one}.
We now show that this is actually not the case. For each $k\in\mathbb{N}$ let
$$\delta_k<\min_{1\leq j\leq k}(t_j-t_{j-1})$$
Let $f_k=\mathbb{1}_{[0,\delta_k]}$. The choice of $\delta_k$ implies that  the intervals $[0,\delta_k]+t_j$, $1\leq j\leq k$ are disjoint. Also, notice that
$ \max_{1\leq j\leq k}f_k(x-t_j)=1$ for $x\in \bigcup^k_{j=1}(\big[0,\delta_k]+t_j\big)$. Thus,
$$ m\big(x\in[0,1]: Mf_k(x)>\tfrac12\big)\geq k\delta_n=\frac{k}{2}\frac{\|f_k\|_1}{\tfrac12}
$$
Hence $\sup_{\lambda>0}\frac{\lambda}{\|f_k\|_1}\,m\big(x\in[0,1]: Mf_k(x)>\lambda)\geq \frac{k}{2}\xrightarrow{k\rightarrow\infty}\infty$. This shows that $M$ does not satisfy \eqref{one}. Consequently, there exists $f\in L_1[0,1]$ for which $m\big(x\in[0,1]:Mf(x)=\infty\big)>0$. For any  $u\in\{x\in[0,1]:Mf(x)=\infty\}$, $f(u-t_n)\stackrel{n}{\nrightarrow}f(u)$.

The rest of this answer is dedicated to sketch a proof of Stein's theorem.
First we state a result similar to Borel Cantelli's lemma:
Lemma 1:
Suppose $(E_n:n\in\mathbb{N})$ is a sequence of measurable sets in $Q=[0,1]^d$ such that $\sum_nm(E_n)=\infty$. Then, there exists translations $F_n=E_n+x_n$ such that
$$\limsup_nF_n=\bigcap_n\bigcup_{m\geq n}F_m =\mathbb{R^d}\qquad \text{a.s.}$$
Idea of proof of Lemma:
This is based in the following geometric. If $A_1,\,A_2\subset Q$, then there is $h\in\mathbb{R}^d$ such that $m\big(A_1\cap(A_2-h)\big)\geq 2^{-d}m(A_1)m(A_2)$.
This follows from the continuity of $g(x):=\big(\mathbb{1}_{A_1}*\mathbb{1}_{A_2}\big)(-x)$ and the fact that $\operatorname{supp}(g)\subset[-2,2]^d=Q'$. For  $g$ attains in maximum in $Q'$ as some point, say $h$, and so
$$m(A_1\cap(A_2-h)=g(h)\geq\frac{1}{m(Q')}\int_{Q'}g(x)\,dx=2^{-d}m(A_1)(A_2)$$
We show that there is a sequence $F_n=E_n+x_n$ of translates such that
$$\bigcup_nF_n=Q\qquad\text{a.s.}$$
The construction is done induction: $F_1=E_1$. If $F_1,\ldots,F_{j-1}$ have been constructed, then let $A_1=Q\cap(F_1\cap\ldots\cap F_{j-1})$ and $A_2=E_j$. Let $h\in\mathbb{R}^d$ be such that $m(A_1\cap(A_2-h))\geq 2^{-d}m(A_1)m(A_2)$. Set $F_j=A_1-h=E_j-h$. Analysis of the proportions $p_j=m(Q\cap(F_1\cap\ldots\cap F_j))$ shows that $p_j-p_{j-1}\geq 2^{-d}(1-p_{j-1})m(E_j)$ from where to follows that $\lim_jp_j=1$.
By decomposing the sequence $E_n$ into countably infinite sub collections, each of which members have measures adding to infinity, and applying the argument above to appropriate translates of the cube $Q$, the conclusion of the Lemma follows.
Sketch of proof of Theorem:
Assume that $\operatorname{supp}(\mu_n)\subset K$. and let $B$ a ball around the origin the contains the support of $\mathbb{1}_Q*\mathbb{1}_K$. Then for any $f\in L_p(\mathbb{R}^d)$ with support in $Q$,  $\operatorname{supp}(f*\mu_n)\subset B$. and the same holds for  $Mf$. The conclusion of the theorem is first proved for $L_p$-functions with support $Q$ (this is enough for the OP).
Suppose \eqref{one} does not hold. Then, for each $k\in\mathbb{N}$ there exist $g_k\in L_p(Q)$ and $\alpha_k>0$ such that
$$ m\big(x\in B: Mg_k(x)>\alpha_k\big)\geq 2^k\frac{\|g_k\|^p_p}{\alpha^p_k}$$
Setting $g'_k=\frac{k}{\alpha_k}g_k$, we have that
$$m\big(x\in B: Mg'_k(x)>k\big)\geq \frac{2^k}{k^p}\|g'_k\|^p_p$$
For each $k$, let $n_k$ the smallest integer such that
$$n_k\frac{2^k}{k^p}\|g'_k\|^p_p\geq1$$
Then  $(n_k-1)\|g'_k\|^p_p <\frac{k^p}{2^k}$ and so
$$\sum_kn_k\|g'_k\|^p_p<\infty$$
This allows us to obtain a sequence $\{f_k\}\subset L_p(\mathbb{R}^d)$ with support in $Q$ (repeating $g'_k$ $n_k$ times) and a sequence $R_k\xrightarrow{k\rightarrow\infty}\infty$ such that
with $E_k=\{x\in B: Mf_k(x)>R_k\}$,

*

*$\sum_km(E_k)=\infty$

*$\sum_k\|f_k\|^p_p <\infty$.

By Lemma 1, there are translates $F_j=E_j+x_j$ such that $\limsup_j F_j=\mathbb{R}^d$ $m$-a.s. Let $\widetilde{f}_j(x)=f(x-x_j)$ and $F(x)=\sup_j\widetilde{f}_k(x)$.  Since the operators $T_n$ are positive ($T_nf\geq0$ whenever $f_n\geq0$), $MF(x)\geq\sup_k M(\widetilde{f}_k)$. Notice that $ M(\widetilde{f}_k)(x)\geq R_k$ for all $F_k$; hence $MF(x)=\infty$ $m$-a.s. On the other hand, $F^p\leq\sum_k\widetilde{f}^p_k$ and so
$$\|F\|^p_p\leq\sum_p\|f_k\|^p_p<\infty$$
This contradicts the assumption that $m(Mf<\infty)>0$ for all $f\in L_p(\mathbb{R})$. Therefore \eqref{one} holds for all $f\in L_p(\mathbb{R}^d)$ supported in $Q$.
To extend the result to any $f\in L_p(\mathbb{R}^d)$ consider translates $Q_j=Q+j$, $j\in\mathbb{Z}^d$ and decompose
$$f=\sum_{j\in\mathbb{Z}^d}f\mathbb{1}_{Q_j}$$
Then $Mf\leq \sum_jMf_j$, and $Mf_j$ is supported in $B_j=B+j$. Any point in $\mathbb{R}^d$ is in at most $N_d$ translates of the form $B+j$ ($N_d$ is a constant depending on $d$ and on the radius of $B$). Applying the result for functions supported on $Q$ we have
$$ m\big(x\in B_j: Mf_j(x)>\lambda\big)\leq\frac{A}{\lambda^p}\|f_j\|^p_p$$
Putting things together,
$$\begin{align} m\big(x: Mf(x)>\lambda\big)&\leq\frac{A N_d}{(\lambda/N_d)^p}\sum_j\|f_j\|^p_p=\frac{AN^{p+1}_d}{\lambda^p}\|f\|^p_p\end{align}$$

More details and generalizations of results of these type can be count in Stein's paper

*

*E. Stein. On limits of sequence of operators, Annals of Mathematics, 1961. Vol. 74, pp. 140-170.

*S. A. Sawyer. Maximal inequalities of weak type, Annals of Mathematics, 1966, Vol 84, pp. 157-174

