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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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You need to notice first that you are decomposing your given function in two others wich are both measurable, because you are taking inverse images of intervals of the form $[0,\infty)$ and $(-\infty,0]$. I'm not sure if I get wat you are asking, so I would like to ask you if you can formulate your question a bit more precisely. – matgaio May 1 '12 at 0:01
@matgaio: Done. – Dan May 1 '12 at 0:08

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,


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


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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Observe that we use strongly the fact that $f$ is integrable (hence measurable) to get the sets $E^+$ and $E^-$ to be measurable and to apply the hypotesis on them – matgaio May 1 '12 at 0:23
Thanks very much for your response. – Dan May 1 '12 at 2:20
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 – matgaio May 1 '12 at 2:29

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