I'm currently tackling the following:
Show, from first principles that if $f(x)=0$ for a.e. $x$, then $\int_{\mathbb{R}} f(x)dm(x)=0$
Now I know how to do this for the other way around (integral equals zero implies that the function equals zero) but am not sure if there is anything else I should do to make sure this way works.
Any help is much appreciated, thanks.
Suppose first that $f\geq 0$ and consider a non-negative simple function $g:x\mapsto\sum_{k=1}^n a_k1_{A_k}(x)$ with $g\leq f$ and $\forall k, a_k\neq 0$. Since $f=0$ a.e, $g=0$ a.e, hence $P(\bigcup_{k=1}^n A_k)=0$, thus $\forall k, P(A_k)=0$, so $$\int g = \sum_{k=1}^n a_k P(A_k)=0$$
Since $f$ is the pointwise increasing limit of simple functions $g_n$ with $g_n\leq f$, the monotone convergence theorem yields $\int f = 0$.
Dropping the non-negativity assumption on $f$, write $f=f^+-f^-$. Since $f=0$ a.e, it follows that $f^+=0$ a.e and $f^-=0$ a.e. Hence $$\int f= \int f^+ - \int f^- = 0-0 =0$$