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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...2010-11-06

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