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Let $f,g $ be non negative measurable function and let $B$ be a measurable set. If $\int_{A} f\, d \mu = \int_{A} g \, d \mu $ for every $ A \subset B$ then $ f= g $ $\mu$-a.e in $B$ (suppose every integral is finite). May someone give me a hint for this one? I tried arguing by contradiction but $f $ and $g$ may have the same integral even on a set of positive measure on which they are not equal..

GaC
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    It might be easier to show that if $\int_A f,\mathrm{d}\mu=0$ then $f=0$ almost everywhere? – SamM Sep 28 '16 at 09:33

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Hint: If $f$ and $g$ are measurable, then the set of $x$ In $B$ for which $f(x)>g(x)$ is measurable.

Ben Grossmann
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To expand on SamM's comment, by linearity of the Lebesgue integral, if $$\int_A (f-g)d\mu =0 \iff \int_A f d\mu - \int_A g d\mu =0 \iff \int_A f d\mu = \int_A g d\mu $$ Thus it suffices to show that $$\int_A f d \mu = 0 \iff f=0 \text{ $\mu-$almost everywhere}$$ since of course $$(f-g)=0\text{ almost everywhere} \iff f=g\text{ $\mu$-almost everywhere}.$$

Chill2Macht
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    Thanks you very much. Now i've got to show that if $\int_{A}f d \mu = 0$ for every $ A \subset B$ then $ f= 0$ $\mu-$almost everywhere on $B$. right? – GaC Sep 28 '16 at 09:52
  • @SpettroDiA Yes exactly. I would also combine this with the other answer. – Chill2Macht Sep 28 '16 at 09:54