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How do I use this the following result

if $f$ is a non-negative measurable function on $X$, then $\int_X f~d\mu =0$ if and only if $f=0$ a.e. on $X.$

to prove that

if $f$ be an integrable function over $X$, then $\int_E f~d\mu =0$ for every measurable subset $E$ of $X$ if and only $f=0$ a.e. on $X$.

In general how does one approach these types of proof where one proves the result for $f\ge 0$ and apply the result to $f^+$ and $f^-$.

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    @matgaio: Done.2012-05-01

1 Answers 1

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You can do it this way:

First, let $E^+=\{x\in X;f(x)\geqslant 0\}$ and $E^-=\{x\in X;f(x)\leqslant 0\}$. Both are measurable, because they are inverse images of intervals by an integrable (hence measurable) function. By hypothesis,

$\int_{E^+}f=0$

Once we have $f\geqslant 0$ on $E^+$, we use the fact mentioned above to get $f=0$ for almost every point on $E^+$. Use the same argument with $-f$, since $-f\geqslant 0$ on the set $E^-$, and we get by hypothesis that

$\int_{E^-}-f=-\int_{E^-}f=0$

and then $-f=0$ for almost every point in $E^-$. Hence $f=0$ for almost every point in $E^-$. Since $X=E^+\cup E^-$, we get the result.

The converse is certainly true as well: if $f=0$, its integral will be zero over each measurable subset of $X$.

I hope this helps you.

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    You are welcome. I hope it was useful. Don't forget to mark it with the green "check" if you think it was useful and correct, to help making the politics of ranking work on math.exchange2012-05-01