$\int_a^b{f(x)dx}=0$ if $f$ is zero a.e.

Question

I am trying to show that if a function $f$ is zero almost everywhere then $\int_a^b{f(x)dx=0}$.

I have shown the inverse implication and have seen other solutions to that, but none with the implication in the forward direction.

For the inverse, I used MCT.

Answer 1

The set where $f$ is non-zero has null measure. Hence the integral on that set equals $0$. It is well know that if you integrate on a set of null measure the integral is always $0$. The integral where $f$ is $0$ is of course $0$. So you can conclude that the total integral is $0$.

Answer 2

$\int f= \int_{f=0} f + \int_{f\neq 0} f =0+0$, where the first term is zero because the integrand is zero everywhere and the second is zero because the set over which you integrate has measure zero.