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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
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1 Answer

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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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 math.exchange –  matgaio May 1 '12 at 2:29
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