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?
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 $f\in L^1(\mu)$.