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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 for Lebesgue almost every $x$, $f_n(x)=f(x-\alpha_n)$ fails to converge to  $f(x)$?

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 $x$.

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 $x$.

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 $f_n(x)=f(x-\alpha_n)$ fails to converge to $f$ Lebesgue almost surely?

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 $x$.

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 for Lebesgue almost every $x$, $f_n(x)=f(x-\alpha_n)$ fails to converge to $f(x)$?

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 $x$.

Suppose $\alpha_n$ is a sequence of positive numbers converging to $0$. Is there a bounded measurable function $f$, say $1$-periodic, such that $f_n(x)=f(x-\alpha_n)$ fails to converge to $f$ Lebesgue almost surely?

This of course is not the case when $f$ is continuous, or when its set of discontinuities has measure zero. If such a function, if 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 $f(x)$ a very sensitive to small variations of the argument $x$.

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 $f_n(x)=f(x-\alpha_n)$ fails to converge to $f$ Lebesgue almost surely?

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 $x$.