Maybe a silly question but can I show that the Lebesgue integral of a function which is zero everywhere is also zero.
I have approached this with a "standard machine" type proof. I can easily show it (1) assuming that the function is a characteristic function, (2) with a simple function, and (3) extending by monotonicity with non-negative function , but I get stuck in (4) any function.
Let's asuume that $h$ is some function. I know by definition that $h=h^+-h^-$ and that the $\mu(h)=\mu(h^+) - \mu(h^-)$, where $\mu(h)=\int h d\mu$ for some measure $\mu$. Now I have that $h=0$, which means that $h^+=h^-$, which implies that the two parts are zero, since $max(f(x),0)=-min(f(x),0)$ only for $f(x)=0$. Now, how do I translate that to the integrals?
Ok, here is the answer thanks to @Yanko using the fact that $h^+=h^-=0$. Then, since I already showed in (3) of the "standard machine" that when a non-negative function is zero, its integral is zero, this means the integrals of the two functions are zero, thus $\mu(h)$ is also zero.