I am trying to prove that if $f$ is an extended-real value function that is measurable, and $f$ is zero almost everywhere, it is then Lebesgue integrable, and $\int f d\mu =0$.
I am struggling with finding a simple proof for this. The issue is that the set where the function is non-zero might be uncountable, so I can't see how to apporximate it with simple functions.
Let $A = \{x : |f(x)| > 0\}$. Then $A$ has measure zero. Any nonnegative nonzero simple function $s$ satisfying $s < |f|$ must be zero everywhere in $A^c$ and therefore must be of the form $$s(x) = \sum_{n=1}^{N} s_n \chi_{S_n}$$ where the $s_n$ are nonzero and $S_n \subseteq A$. Therefore, each $S_n$ has measure zero, and so $$ \int s(x)\ dx = \sum_{n=1}^{N} s_n \int_{S_n} \ dx = \sum_{n=1}^{N} s_n\ m(S_n) = 0$$ Since $\int |f|$ is the supremum of all integrals of nonnegative simple functions satisfying $s < |f|$, we have $\int |f| = 0$, which implies that $\int f = 0$.