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I'm trying to solve the following problem from Klenke (3rd ed, exercise 4.2.1.)

Let $(\Omega, \mathcal{A},\mu)$ be a measure space and let $f\in L^1(\mu)$. Show that for any $\epsilon>0$, there is an $A\in\mathcal{A}$ with $\mu(A)<\infty$ and $|\int_{A}fd\mu-\int fd\mu|<\varepsilon$.

I can show this if I can assume that $\mu$ is finite, i.e., $\mu(\Omega)<\infty$, by using the continuity of Lebesgue integral. Specifically,

$$ |\int_{A}fd\mu-\int fd\mu| =|\int_{\Omega}fd\mu-\int_{\Omega\setminus A}fd\mu-\int fd\mu| =|\int_{\Omega\setminus A}fd\mu| \le \int_{\Omega\setminus A}|f|d\mu, $$

so I can take $B:=\Omega\setminus A$ such that $\mu(B)<\delta$ and the RHS of the above is less than $\varepsilon$. Then, $\mu(A)=\mu(\Omega)-\delta$, so if $\mu(\Omega)<\infty$, we have $\mu(A)<\infty$.

But, how can I show this when $\mu$ is not finite? Can I somehow generalize the arguemnt above?

keepfrog
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    Use DCT on $|f| \chi_{{ |f| \geq n}}$. – FZan Jun 27 '23 at 19:22
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    If you assume $\mu$ to be $\sigma$-finite then you have $A_{n}$ disjoint such that $\bigcup_{n\in\Bbb{N}}A_{n}=\Omega$ with $\mu(A_{k})<\infty$ . So $\displaystyle f\mathbf{1}{\bigcup{k=1}^{n}A_{k}}\xrightarrow{n\to\infty}f$ . Now you can use DCT to find $N$ such that for all $n\geq N$, you have $\bigg|\displaystyle\int_{\Omega}f-\int_{\Omega}f\mathbf{1}{\bigcup{k=1}^{n}A_{k}}\bigg|<\epsilon$ for all $n\geq N$ and in particular you can take $\displaystyle A=\bigcup_{k=1}^{N}A_{k}$. – Mr. Gandalf Sauron Jun 27 '23 at 19:43
  • @MichaelGreinecker Ofcourse. Having a complete brain freeze today. – Mr. Gandalf Sauron Jun 27 '23 at 20:34
  • Why is the function note integrable when $\mu$ is not $\sigma$-finite? – keepfrog Jun 27 '23 at 20:42
  • @keepfrog ignore michaels's comment. It was in reply to one of my comments which I deleted. – Mr. Gandalf Sauron Jun 27 '23 at 21:11
  • @Mr.GandalfSauron Does it mean your answer is not complete because it does not cover the case where $\mu$ is $\sigma$-finite? Or, as Michael's comment suggests, if it is not $\sigma$-finite then $f\notin L^1$? In that case, why can we conclude that $f\notin L^1$? – keepfrog Jun 27 '23 at 22:03
  • @FZan I see that I can apply the DCT to $f \mathbb{I}_{{|f|>1/n}}$ in a similar manner as Mr.Gandalf's comment. But in the case, how can we ensure that $\mu({|f|>1/n})<\infty$ for some large $n$? – keepfrog Jun 27 '23 at 22:06
  • Well you do know $\int |f|$ is finite, try using Chebiyev. – FZan Jun 27 '23 at 22:35
  • @FZan Ah I see thank you – keepfrog Jun 27 '23 at 23:01
  • @keepfrog The comment I gave answers the question for $\mu$ being $\sigma$-finite. Additionally I added a comment that what does FZan's comment achieve since we cannot know whether ${|f|\geq n}$ has finite measure. To which Michael pointed out that if those sets do not have finite measure then $|f|$ cannot possibly be integrable. So as I said, as of now, ignore Michel's comment. – Mr. Gandalf Sauron Jun 28 '23 at 07:38
  • But still, @FZan 's comment does not help us see whether ${|f|\leq n}$ has finite measure.(Which is what is required by FZan's method as you would have to set $A={|f|\leq N}$ to get $\int_{|f|\geq N}|f|<\epsilon$. For example, when $f$ is identically $0$, then all such ${|f|\leq n}$ have infinite measure(if $\mu$ is not finite) but still $f$ is integrable. So I don't really see how FZan's comment would help there. – Mr. Gandalf Sauron Jun 28 '23 at 07:43
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    @Mr.GandalfSauron This is what I had in mind: If we apply the DCT to $f\mathbb{I}{{|f|>1/n}}$ instead, then for large $n$ we have $|\int f\mathbb{I}{{|f|>1/n}} d\mu - \int f d\mu|<\epsilon$. Thus, we can take $A={|f|>1/n}$, and by Chebyshev $\mu(A)\le n \int |f| d\mu < \infty$. – keepfrog Jun 28 '23 at 20:49
  • @keepfrog It seems good to me. You can post it as an answer yourself. – Mr. Gandalf Sauron Jun 29 '23 at 07:26

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I have figured out an answer with the community's help in the comment section, so I post that for completeness:

We can apply the DCT to $f\mathbb{I}_{\{|f|>1/n\}}$, so we have $\lim_{n\to\infty}\int f\mathbb{I}_{\{|f|>1/n\}}d\mu=\int fd\mu$. This implies that, for large $n$, we have $|\int f\mathbb{I}_{\{|f|>1/n\}}d\mu-\int fd\mu|<\epsilon$. Therefore, we can take $A=\{|f|>1/n\}$. Finally, by Chebyshev's inequality, $$ \mu(A)=\mu(\{|f|>1/n\})\le n\int|f|d\mu<\infty, $$ where the last inequality is from $f\in L^1(\mu)$.

keepfrog
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