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PROBLEM: 1) Assume $f$ is a bounded measurable function on $\mathbb{R}$, does $f_n(x):=f(x+\frac{1}{n})$ converge to $f(x)$ almost everywhere? 2) If not, is there a subsequence $n_k$ such that $f_{n_k}$ converges to $f$ almost everywhere?

Beginner in Real Analysis and have two ideas but neither worked:

  1. Lusin's Theorem. Not sure how the statement would be if the domain of $f$ is $\mathbb{R}$. One related problem here: Version of Lusin's theorem on real line.
  2. Dirichlet Function as an example. When $x$ is irrational, whether $f_n(x)$ converges to $f(x)$ or not.

Any hints or counterexamples will be appreciated.

Maul Seil
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  • What happens if $f$ has a (right) discontinuity? – Henry Oct 25 '20 at 17:20
  • @GiuseppeNegro I suppose bounded and measurable has tiny difference but quite similar. Thank you! – Maul Seil Oct 25 '20 at 17:32
  • There is no difference. Indeed, the counterexamples that you find in the linked page are all of the form $\chi_K$ for a compact $K$. (And $\chi_K$ is the function that is $1$ on $K$ and zero everywhere else). There are basically two counterexamples on that page: one is probabilistic (David C. Ullrich, myself) and the other is deterministic (Alex Ravski, Dap). – Giuseppe Negro Oct 25 '20 at 17:35
  • @GiuseppeNegro bounded and measurable is not equivalent to $L^1$ right? I'm asking this because if so is there still a subsequence $n_k$ such that $f_{n_k}$ converges to $f$ almost everywhere based on bouded and measurable but not $L^1$? – Maul Seil Oct 25 '20 at 18:11
  • No, of course "bounded and measurable" is not equivalent to $L^1$. I am not sure I understand your question about subsequences. What I meant above is that you can find a bounded and measurable function $f$ such that $f(\cdot + \frac1n)$ does not converge on a set of nonzero measure. – Giuseppe Negro Oct 25 '20 at 22:39
  • @GiuseppeNegro The original problem has two parts. I got stuck in the first one so I did not mention the second one in the beginning. PROBLEM: Assume f is a bounded measurable function on R, does fn(x):=f(x+1/n) converge to f(x) almost everywhere? If not, is there a subsequence $n_k$ such that $f_{n_k}$ converges to $f$ almost everywhere? For the first, I believe the linked page already gave detailed counterexamples. For the second, since not $L^1$ (stated at the beginning in the linked page and the existence of such subsequence), existence is not immediate? – Maul Seil Oct 26 '20 at 01:40
  • @GiuseppeNegro (Comment is limited) As I understand, thanks to the linked page, the second part is similar to prove there are infinite $n_k$ such that $x+1/{n_k} \in K$ for a compact $K$ is the counterexample for the first part ($f$ should be the indicator function of $K$). Is that right? – Maul Seil Oct 26 '20 at 01:49
  • @user567445: the question is indeed interesting. I don't remember the details of those counterexamples and I can't tell you straight away if they admit a convergent subsequence or not. I would say they don't, but that's just an educated guess. – Giuseppe Negro Oct 26 '20 at 19:54
  • Why didn't you ask about convergent subsequences in the question? The way you phrased it, it is a duplicate of the other one. So it got closed. But the subsequence problem is novel and interesting. If you edit the question, including the subsequence problem, I will vote to reopen it. Maybe some of the people involved in the other question will come here and tell us what they think. – Giuseppe Negro Oct 26 '20 at 22:19
  • I have a solution to the subsequence problem. It is actually quite easy. I claim that there is a subsequence $(\tfrac{1}{n_k})$ such that $f(x+\tfrac1{n_k})\to f(x)$ for almost all $x$. The only assumption is that $f\in L^1(\mathbb R)$.

    Proof. It is a standard property that $f(\cdot + \frac1n)\to f$ in $L^1$ sense (this is called "$L^1$ continuity of the translations"). It is another standard property that a sequence that converges in $L^1$ has a subsequence converging pointwise and almost everywhere. This concludes the proof.

     – Giuseppe Negro Oct 27 '20 at 13:31
  • @GiuseppeNegro We don't have any assumption for $f\in L^1(\mathbb R)$ at the beginning. You lost me there. – Maul Seil Oct 27 '20 at 14:05
  • Right. Just sloppiness on my part, nothing substantial. So our $f$ is bounded and measurable. Ok, so let us fix an interval $[a,b]$. Now clearly $f\in L^1(a, b)$, being bounded and measurable. So we can apply our reasoning here, on this interval, and we find a subsequence $1/n_k$ such that $f(x+\tfrac1{n_k})\to f(x)$ for almost all $x\in[a, b]$. Is it clear until now? – Giuseppe Negro Oct 27 '20 at 14:11
  • @GiuseppeNegro I fully understand your last two comments. But when it comes to real line I’m not sure this is enough. – Maul Seil Oct 27 '20 at 14:30
  • Of course it is not. For that, we need a diagonal sequence argument. Let’s do $[0, \infty)$ instead of $\mathbb R$. Consider the interval $[0, 1]$. We find a subsequence $1/n_1(k)$. (I use parenthesis instead of pedices). Now let us pass this subsequence on to $[1, 2]$. We find $1/n_2(k)$, which is a subsequence of $1/n_1(k)$. And so on. In the end, the sequence we take is $1/n_{k}(k)$, the diagonal one. (This is a classical argument in analysis, called “diagonal sequence trick” or similar names). By construction we have that $f(x+\tfrac1{n_k(k)})$ converges ae [...] – Giuseppe Negro Oct 27 '20 at 14:40
  • [...] on all intervals $[0, 1], [2, 3], [3, 4] $ and so on. And so this sequence converges ae on $[0, \infty)$. – Giuseppe Negro Oct 27 '20 at 14:41
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    @GiuseppeNegro Fruitful ideas! Thanks! – Maul Seil Oct 27 '20 at 14:50

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