Does $f\Big(x+\frac{1}{n}\Big) \to f(x)$ for a.e. x as $n \to \infty$?

Question

Let $f$ be a bounded measurable function on the real line, then is it true that $f\Big(x+\frac{1}{n}\Big) \to f(x)$ for a.e. $x$ as $n \to \infty$

I found a result where this is true for $f\in L^1(\Bbb{R})$, and I think in general it may not be true, but I'm not able to find a counter.

Answer 1

I would guess that question may come from Lusin's theorem.

It states that for every measurable function there exists measurable (non-zero measure) subset where it is is continuous. So function is not necessary continuous on its entire set. Let $$f(x) = 1_{{sqrt(x)\in Q}}$$ and $0$ otherwise.

So $f(x)$ is bounded and measurable but not continuous in rational points that are full squares.

I am not sure that a.e. states for. If it is "almost everywhere", i.e. set has to have non-zero measure then my example is not valid.

UPDATE

I have looked up on Lusin theorem variants. Proposition as it stated is current post not true... However, the following statement is true:

If $f \in L^1(\lambda)$ and $\epsilon > 0$. Then there is a continuous function $g \in L^1(\lambda)$ so that $$\int_{R} |f - g| d\lambda \lt \epsilon$$ I.e. measurable function from $L^1$ can be approximated by continuous function from $L^1$. It is not point-wise convergence, but weak convergence on measure. See "7 Theorem" here.

Answer 2

The answer is NO. We construct an example when $f(x+1/n)\not\to f(x)$ on a set with positive measure.

Set up a variable base number system: for $0\le x<1$, let $$ x=\sum_{k=1}^\infty \frac{a_k(x)}{10^{1+2+\ldots+k}} \quad\text{where $0\le a_k(x)<10^k$}, $$ and let $$ f(x) = \begin{cases} 1 & \exists k ~ a_k(x)=0 \\ 0 & \text{otherwise.} \\ \end{cases} $$

Let $A=\big\{ x\in[0,1): \forall k ~ a_k(x)\ge2 \big\}$; that is the set of those numbers that do not contain the digits $0$ and $1$ at all. This set has positive measure: $\lambda(A)=\prod\limits_{k=1}^\infty \left(1-\frac2{10^k}\right)>0$. We show that for every $x\in A$, we have $f(x)=0$ and $f(x+1/n)=1$ for infinitely many $n$.

It is trivial that $f(x)=0$ for $x\in A$.

Take an arbitrary $x\in A$; We need infinitely many positive integers $n$ such that $f(x)=1$. Take an integer $k>3$; we want to change the $k$th digit to $0$ or $1$ by adding $1/n$ to $x$. Of course we can write $1/n$ in our number system; we can see that $$ a_k(x+1/n) = a_k(x)+a_k(1/n) \quad\text{or}\quad a_k(x+1/n) = a_k(x)+a_k(1/n)+1, $$ depending the carry from the lower significant digits. Hence, it suffices to enforce $a_k(x)+a_k(1/n)=10^k$.

Let $b=10^k-a_k(x)$ and take an integer $n$ such that $$ a_k(1/n)=b, \quad\text{i.e.}\quad \frac{b}{10^{1+2+\ldots+k}} \le \frac1n < \frac{b+1}{10^{1+2+\ldots+k}}, $$ or equivalently $$ \frac{10^{1+2+\ldots+k}}{b+1} < n \le \frac{10^{1+2+\ldots+k}}{b}. $$ Such an $n$ exists if $$ \frac{10^{1+2+\ldots+k}}{b}-\frac{10^{1+2+\ldots+k}}{b+1} > 1; $$ this is true because $$ b(b+1) < 10^{2k} < 10^{1+2+\ldots+k}. $$ Therefore the desired $n$ exists. Since $n>10^{1+\dots+(k-1)}$, the set of the constructed values for $n$ is unbounded.

So, $f(x+1/n)\not\to f(x)$ in the set $A$ with positive measure.