Almost surely convergence of translations of a measurable function Suppose that $\alpha_n$ is a sequence of positive numbers converging to $0$.

Question. Is there a bounded measurable function $f$, say $1$-periodic, such that  $f_n(x)=f(x-\alpha_n)$ does not converge to $f(x)$ for any $x$ in  some set $A$ of positive Lebesgue measure?

This of course is not the case when $f$ is continuous, or when its  set of discontinuities has measure zero. In fact such a function, if it exists, must have support on a totally disconnected set of the real line.
Edit: If such $f$ exist, then considered as a function on $\mathbb{T}$ one would have
$$ \|f_n-f\|_{L_1(\mathbb{T})}\xrightarrow{n\rightarrow\infty}0$$
and so, every subsequence fo $f_n$ would have an almost surely convergent subsequence; which makes $f(x)$  very sensitive to small variations of the argument $x$.
 A: These questions boil down to questions about maximal inequalities. If you define $(T_nf)(x)=f(x-\alpha_n)$, then maximal inequalities are concerned with the operator $Mf(x)=\sup_n |T_nf(x)|$ and the finite approximations $M_nf(x)=\max_{k\le n}|T_kf(x)|$.
One then asks whether the (non-linear) operator $M$ is bounded in the "weak $L^1$ norm" (which is not a norm at all!). The weak $L^1$ norm of a function $f$ is $\|f\|_{1,\infty}:=\sup_t t\cdot\lambda\{x\colon |f(x)|\ge t\}$ (that is the area of the largest rectangle under the graph). A key example of something with finite weak $L^1$ norm, but infinite $L^1$ norm is $f(x)=\frac 1x$ (which clearly has weak $L^1$ norm 1).
A weak $L^1$ maximal inequality is an inequality of the form $\|Mf\|_{1,\infty}\le C\|f\|_1$ for all $f\in L^1$. One can show (as in this question) that if a weak $L^1$ maximal inequality is satisfied, then the set of $f$ for which $T_nf$ converges pointwise to $f$ is an $L^1$ closed set. Since for continuous functions $T_nf$ converges pointwise to $f$, and continuous functions form a dense set, if a weak $L^1$ maximal inequality is satisfied, it follows that $T_nf$ converges pointwise to $f$ for all $f\in L^1$.
The converse is also true: if there is not a weak $L^1$ maximal inequality, then there are $L^1$ functions for which $T_nf$ does not converge pointwise to $f$. This is based on (for example) taking positive $f_j$'s of $L^1$ norm $2^{-j}$ so that $Mf_j$ is of weak $L^1$ at least norm $2^j$. Summing the $f_j$'s gives a counterexample.
In this case, it is not hard to see that there is no weak $L^1$ maximal inequality: suppose that the $(\alpha_n)$ are strictly decreasing (this is not essential, but simplifies things). Let $N\in\mathbb N$ and let $\delta<\min_{i\le N}(\alpha_i-\alpha_{i+1})$. Then if $f$ is the indicator function of an interval $J$ of length $\delta$, then $M_Nf$ takes the value 1 on $\bigcup_{i=1}^N(J+\alpha_i)$. By the choice of $\delta$, these intervals are disjoint, so we see that $\|Mf\|_{1,\infty}\ge N\|f\|_1$ for this $f$.
In your question, you asked for a bounded function. You can do this in the same way, but you have to tinker with the construction by hand to make sure it does the right thing.
Edited to add details of (unbounded) $L^1$ counterexample
You can pick out a subsequence of the $\alpha_n$'s so that $\alpha_n<\frac 12\alpha_{n-1}$. Clearly if you don't have pointwise convergence along this sequence, nor do you have pointwise convergence along the full sequence.
Let $k$ be given. If $f$ is the indicator function of an interval of length $\alpha_{4^{k+1}}$, then $T_{4^k}f,\ldots, T_{4^{k+1}-1}f$ are disjointly supported. $\|\max_{4^k\le j< 4^{k+1}}T_jf\|_{1,\infty}=3\cdot 4^k\|f\|_{1,\infty}$.
Now if you take $g_k$ to be a randomly chosen sum of $3^{-k}/\alpha_{4^{k+1}}$ translates of $f$, you obtain a function of norm $3^{-k}$ where for each $x$, the probability that $\max_{4^k\le j< 4^{k+1}}T_jg_k$ is not 1 is $(1-3^{-k})^{3\cdot 4^k}$, which is microscopic. Hence $g_k$ is very likely to give you "hits" when you apply $T_j$'s in the range $4^k$ to $4^{k+1}$. Summing the $g_k$'s, you get a function giving hits in all of these ranges.
