Show that: $\int_{\Omega} X d\mu = 0 \iff \mu(\{\omega\in\Omega\mid X(w)>0\})=0$

Question

Assignment:

Let $(\Omega,\mathfrak{A},\mu)$ be a measure space and $X: \Omega \rightarrow \bar{\mathbb{R}}$ a non-negative $\mathfrak{A}$-$\bar{\mathfrak{B}}$-measurable function. Show that: $$\int_{\Omega} X d\mu = 0 \iff \mu(\{\omega\in\Omega\mid X(w)>0\})=0$$

This should be a simple proof (I believe), but I'm struggling. My approach (using non-negativeness of $X$ and a disjunct partition): $$0 = \int_{\Omega} X d\mu = \int_{\{\omega\in\Omega: \ X(\omega) >0\}\uplus \{\omega\in\Omega: \ X(\omega) ≤ 0\}}X d\mu=\int_{X(\omega) > 0}X d\mu + \int_{X(\omega) ≤ 0}X d\mu = \int_{X(\omega)>0} Xd\mu = \int X \cdot \chi_{X(\omega)>0} d\mu$$ But the right-hand side of what I need to show is: $$\mu(\{\omega\in\Omega\mid X(w)>0\}) =\int\chi_{X(\omega)>0} \neq \int X \cdot \chi_{X(\omega)>0} d\mu$$

So I'm (a) wondering where my mistake is and (b) what an alternative (better) approach would be (e.g. one where I actually so both implications separately and not simultaneously like I am trying to).

Answer 1

"$\Leftarrow$" is easy. For "$\Rightarrow$", assume the contrary. Since

$$\{\omega\in\Omega:X(\omega)>0\}=\bigcup_{n=1}^{\infty}\{\omega\in\Omega:X(\omega)>\frac{1}{n}\}$$

Then there is at least one $N\in\mathbb{N}$ s.t.

$$\mu(\{\omega\in\Omega:X(\omega)>\frac{1}{N}\})>0$$

Then conclude that

$$\int_\Omega Xd\mu\geq\int_{\{\omega\in\Omega:X(\omega)>\frac{1}{N}\}}Xd\mu>\int_{\{\omega\in\Omega:X(\omega)>\frac{1}{N}\}}\frac{1}{N}d\mu=\frac{1}{N}\mu(\{\omega\in\Omega:X(\omega)>\frac{1}{N}\})>0$$

which is a contradiction

Answer 2

Motto: Compare expectations... Since $X\geqslant0$ almost surely, $nX\geqslant\mathbf 1_{nX\geqslant1}$ almost surely, hence $$0=n\int_\Omega X\mathrm d\mu=\int_\Omega nX\mathrm d\mu\geqslant\mu(nX\geqslant1)\geqslant0,$$ thus, $$\mu(nX\geqslant1)=0.$$ To conclude, note that $$\mu(X\gt0)=\bigcup_{n\geqslant1}\mu(nX\geqslant1).$$