I have a positive, $\sigma$-finite measure $\nu$ and a function $h$. I have the integral $\int h \,d\nu$. Now I'm trying to show that if $h=0$, then the Lebesgue integral is also zero. I try to use the standard machine technology but I'm not quite sure if my reasoning in the first step is correct.
Let's assume that $h$ is a characteristic function, such that $h(x)=1$ if $x \in A$ and $x=0$ if $x \in A^c$. Since $h=0$ for any $x,$ then $x \in A^c$. Then $\int h \,d\nu = \nu(A^c)=0$. I guess this is wrong though, since I cannot be sure that $ \nu(A^c)=0$ and I'm not even sure if I can write this that way. Any hints then how I should approach this?
You can use linearity of integral: $h(t)\equiv 0$, then $h(t)=\chi_E(t)-\chi_E(t)$ for some (any) measurable set $E$.
$$ \int h\,d\nu = \int\left[\chi_E(t)-\chi_E(t)\right]\,d\nu = \int\chi_E\,d\nu-\int\chi_E\,d\nu = \nu(E)-\nu(E)=0 $$
Or you can use the characteristic function of the empty set: $h(t)\equiv \chi_\emptyset(t)$.
$$ \int h\,d\nu = \int\chi_\emptyset (t)\,d\nu = \nu(\emptyset)=0 $$
They're basically the same.