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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1It cannot be. If $A$ is a non measurable set in $[0,1]$, then define $f = -1 +2(1_A)$. Then $|f|=1$ and isintegrable, but $f$ is not. – copper.hat Oct 08 '12 at 01:02
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2Maybe you are looking to add the hypothesis that $f$ is measurable function into the (possibly extended) reals? – Oct 08 '12 at 01:14
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4Look at $f^+$ and $f^-$ – leo Oct 08 '12 at 02:50
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@leo's comment hold the key... – copper.hat Oct 08 '12 at 03:03
2 Answers
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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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