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I want to prove that a measurable function $f$ is Lebesgue integrable iff $|f|$ is. I've proved the first part but how can I show if $|f|$ is Lebesgue integrable then $f$ is ?

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    @leo's comment hold the key...2012-10-08

2 Answers 2

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Well the both parts are easy.

If $f$ is Lebesgue integrable than $f^+$ and $f^-$ are Lebesgue integrable, so integral of both $f^+$ and $f^-$ is finite. Thus, their sum is also finite, which means, integral of $|f|$ is finite. If $f$ is measurable than $|f|$ is measurable. If you combine last 2 statements, you get that $|f|$ is Lebesgue integrable.

Same in the other direction, if $|f|$ is Lebesgue integrable, $f^+$ and $f^-$ are also, so integral of $f^+$ and integral of $f^-$ are finite, thus their subtraction is finite, which is integral of $f$.

$f^+ = \max\{0,f\}$

$f^- = \max\{0,-f\}$

$f = f^+ - f^-$

$|f|= f^+ + f^-$

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This is not true. Let $M$ be a non-measurable "monster" set contained in $[0,1]$ Define $f(x) = 1$ if $x\in M$ and $-1$ otherwise. We see that $|f| = 1$ on $[0,1]$ so $|f|$ is Lebesgue integrable. Since $M$ is not measurable, $f$ is not.

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    I'm considering measurable functions.2012-10-08