This is problem9 page79 From Royden 4Ed.
Let $E$ have measure zero. Show that if $f$ is a bounded function on $E$, then $f$ is measurable and $\int_{E} f=0$.
My partial solution:-
We can easily shown that $f$ is measurable on $E$ since our measure space is complete and any set $ E_0=\{x\in E : f(x)>c\} \subset E$ hence $E_0$ is measurable implies $f$ is measurable . for the second part I have this:-
$\int_E f dx = \int f \chi_{E}(x) dx \le M \,m(E)=0$. Hence
$$\int_E f dx \le 0$$
but I am stuck in proving the other side of equality.any help to prove
$$\int_E f dx \ge 0$$
Edit:- I think in case $f(x)\ge 0$ monotonicity will give us the other part of inequality for free. and in case $f(x)\le 0$ we get $ -f(x)\ge 0$ by part one we get what we are looking for. is this right?
Recall that $\int_E f = \int_E f^+ - \int_E f^-$ where $f^+=\max\{f,0\}$ and $f^-=\max\{-f,0\}$. You can use your argument to show that both the positive and negative integrals are zero.