Suppose that $\alpha\_n$ is a sequence of positive numbers converging to $0$.



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> **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?
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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 $x$.



Almost surely convergence of translations of a measurable function
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](https://mathoverflow.net/questions/28997/does-anyone-know-an-intuitive-proof-of-the-birkhoff-ergodic-theorem/45250#45250)) 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.


