$\int_{A} f\, d \mu = \int_{A} g \, d \mu $ for every $ A \subset B$ then $ f= g $ $\mu$ a.e in $B$

Question

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..

Answer 1

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 $$(f-g)=0\text{ almost everywhere} \iff f=g\text{ $\mu$-almost everywhere}.$$

Answer 2

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