TITLE: Measurable set on which a function is bounded
QUESTION [5 upvotes]: Let $f$ be in $L^{1}(\mathbb{R})$, $\mathbb{R}$ the real numbers. Show that for every $\varepsilon > 0$ there exists $A \subseteq R$ , measurable, such that $m(A) < \infty$ , $f$ is bounded on $A$ and $ \int_{\mathbb{R}} |f|  < \int_{A} |f| + \varepsilon$.
If we take $A$ as the support of the simple function which approximates $f$ in the $L^{1}$ norm then this has finite measure and it satisfies the other conditions. But I don't see why $f$ must be bounded on it. Any ideas?
Thank you.

REPLY [7 votes]: You can do this as a "$2\epsilon$-proof" (or $\epsilon/2$ if you prefer).  First, since $(\int_{-n}^n|f|)_{n=1}^\infty$ converges to $\int_{\mathbb{R}}|f|$, there is an $n$ such that $\int_{\mathbb{R}}|f|\lt\int_{-n}^n|f|+\epsilon/2$.  Then, since $(\int_{-n}^n|f|\cdot\chi_{\{|f|\leq m\}})_{m=1}^\infty$ converges to $\int_{-n}^n|f|$, there is an $m$ such that $\int_{-n}^n|f|\lt\int_{-n}^n|f|\cdot\chi_{\{|f|\leq m\}}+\epsilon/2$.  Take $A=[-n,n]\cap\{x:|f(x)|\leq m\}$. 
Rather than using simple functions to show this, I would use this as a first step to showing that $f$ can be approximated by simple functions, because now $f$ can be uniformly approximated by simple functions on $A$.

REPLY [3 votes]: $|f|I_{|f|\leq n} $ increases to $|f|I_{|f| < \infty} = |f|$ (a.e.) as $f$ is in $L^{1}(\mathcal{R})$, the result follows from monotone convergence theorem.
Update : This answer is incomplete, please see Jonas's answer.