Is it true that if $f: X \rightarrow \mathbb R$ is integrable on a measure space $(X, M, \mu)$ then for arbitrary $\varepsilon >0$ there exists a set $A \in M$ with finite measure such that $\int_{X\setminus A} |f| d \mu <\varepsilon$?

Edit.

Thanks for help. If I understood correctly, for each $n \in \mathbb N$ the set $A_n:=\{x\in X: |f(x)|>\frac{1}{n} \}$ is of finite measure, the sequence $(X\setminus A_n)$ is decreasing, its intersection is equal $\{x: f(x)=0\}$, $\lim_{n \rightarrow \infty} \int_{X\setminus A_n} |f| d \mu= \int_{ \bigcap_{n\in N}(X \setminus A_n) } |f| d \mu= \int_{ \{x: f(x)=0 \} } |f| d \mu=0$. Hence for given $\varepsilon >0$ there exists $m\in N$ such that $\int_{X\setminus A_m} |f| d\mu < \varepsilon$, and $\mu(A_m)<\infty$.

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I would only add that to make the limit work you want the Lebesgue Dominated (or Monotone) Convergence Theorem. – Robert Israel Feb 8 '13 at 23:01

Hint: let $A_n = \{x: |f(x)| > 1/n\}$.