Let $(X,\mathcal{S},\mu)$ be a measure space and $f:X\rightarrow [0,\infty]$ an $\mathcal{S}$-measurable function. Show that $\int f \, d\mu>0$ if and only if $\mu(\{x\in X: f(x)>0\})>0.$
My thoughts:
For the forward direction, we know there exists some partition $P$ such that:
$\mathcal{L}(f,P)=\mu(A_1) \inf(f)+\cdots+\mu(A_{m})\inf(f)>0.$ So if we consider $x$'s where $f(x)>0$, wouldn't that mean that the infimum of $f$ on all of these $A_i$'s is positive? Then $\mu(A_{1})+\cdots+\mu(A_{m})=\mu(X)>0$? I'm not sure that I'm thinking about this correctly. Any suggestions would be appreciated. Thank you
\begin{align} & \{x\in X: f(x)>0\} \\[8pt] = {} & \bigcup_{n\,\in\,\mathbb N} \left\{x\in X: \tfrac 1 {n+1} \le f(x)< \tfrac 1 n \right\} \cup \{x\in X: f(x)\ge1\}. \end{align} The measure of this union is positive only if the measure of one of the terms is positive. So you need to show that implies the integral is positive.