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Let $f:\mathbb{R}\rightarrow\bar{\mathbb{R}}$ be a Lebesgue integrable function. How to prove that $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}f(x-\sqrt{n})$ converges almost everywhere on $\mathbb{R}$?

I tried to integrate $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}f(x-\sqrt{n})$ on a suitable measurable set to solve the problem just like Prove the series converges almost everywhere but it seems hard for this problem to deduce that the integration of $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}f(x-\sqrt{n})$ on a measurable set is less than $c\|f\|_1$ with $c>0$.

L'enfer
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    Could you please edit your question to elaborate on your overall strategy? It would be great if you could explain the reason why you decided to approach this problem via investigating integrals. What do you hope to prove about the integral of a function that will help you prove that $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}f(x-\sqrt{n})$ converges almost everywhere? Also, you mention "suitable measurable sets", so could you please tell us which sets these are? And of course, please share any observations you've made about any integrals you've investigated so far. – Kenny Wong Apr 10 '23 at 12:34
  • Okay, I see you've edited your question. The approach looks sensible, but it looks like you've given up before you've even begun. Take $E := [0, 1)$, and try your very hardest to prove that $\int_E \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} |f(x-\sqrt{n})| < c \left| f\right|_1$ for some $c > 0$. You don't need sophisticated measure theory to do this. Treat it almost like a high school problem. If you're still stuck after having tried very hard, then perhaps you could edit your post again to show everything you've done. – Kenny Wong Apr 10 '23 at 13:00
  • Thank you very much! I have got the point. However not long ago I find that there was a same post here 5 years ago, see https://math.stackexchange.com/questions/2402988/convergence-of-a-series-of-translations-of-a-lebesgue-integrable-function?rq=1, and the solution in this post is similar to your hint for me. – L'enfer Apr 10 '23 at 14:31
  • Glad to know that you've resolved this! :) – Kenny Wong Apr 10 '23 at 14:40
  • Since this is a duplicate, the right thing to do is to mark it as one. Well done for finding it. – Kenny Wong Apr 10 '23 at 14:41

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