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A problem I have been thinking about but not sure if it is right or not.

Claim: Let $(X, \mathcal{M}, \mu)$ be a measure space and let $f,g$ be real-valued measurable, $L^1$ functions. If $\int_A f\,d\mu = \int_A g\,d\mu$ for all $A\in \mathcal{M}$, then $\int_A f^+\,d\mu = \int_A g^+\,d\mu$ and $\int_A f^-\,d\mu = \int_A g^-\,d\mu$ for all $A\in\mathcal{M}$.

Would appreciate a quick proof or a counterexample, but a reference is fine too.

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    From $\int_A f\,d\mu = \int_A g\,d\mu$ for all $A\in \mathcal{M}$ it follows that $f=g$ almost everywhere, so the positive and negative parts are the same. – 2017-02-03

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Take $A=\{x:f(x)>g(x)\}$. Then $\int_A f\,d\mu=\int_A g\,d\mu$ implies $A$ has measure $0$ (perhaps more straightforwardly seen by considering $h=f-g$). Similarly, $\{x:f(x)

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    Thanks! As a follow-up, does a.e. equality also hold if $\int_A f\,d\mu = \int_A f\,d\mu$ for all $A\in\mathcal{E}$, where $\mathcal{E}$ generates the $\sigma$-algebra $\mathcal{M}$? – 2017-02-04
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    That would follow if you can show that the set of $A$ where equality holds is a sigma algebra. – 2017-02-04