# Measurable set on which a function is bounded

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.

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@user100: take a sequence of functions converging pointwise to $f$ which are bounded (e.g., $f_n$ equal to $f$ if the values are between $-n$ and $n$, and either $-n$ or $n$ otherwise), then approximate $f$ with some bounded function and the bounded function with a simple function. I haven't worked out the details, so this may very well crash and burn... –  Arturo Magidin Nov 6 '10 at 20:03

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 $A$.

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(The 2 steps aren't really necessary, and you could just use the fact that $|f|\cdot \chi_{\{x\in[-n,n]:|f(x)|\leq n\}}$ converges to $|f|$ pointwise a.e.) –  Jonas Meyer Nov 6 '10 at 20:23
Nice, thank you Jonas. –  student Nov 7 '10 at 1:05
@user100: If you like it you should approve the answer :) –  AD. Nov 7 '10 at 13:05
@user100: You're welcome. Of course this is just a sketch, and justification is needed for the 2 claims of convergence. For example, monotone or dominated convergence can be used. –  Jonas Meyer Nov 7 '10 at 18:04

$|f|I_{|f|\leq n}$ increases to $|f|I_{|f| < \infty} = |f|$ (a.e.) as $f$ is in $L^{1}(\mathcal{R})$, the result follows from monotone convergence theorem.

Sorry for the overlap. I was still getting the TeX to work when your answer was posted. Your answer would have to be modified to get $m(A)\lt\infty$. –  Jonas Meyer Nov 6 '10 at 20:38