Prove that $\int f\,d\mu>0 \text{ if and only if } \mu(\{x\in X: f(x)>0\})>0.$

Question

I am having a particularly difficult time starting the following problem

Suppose $(X,\mathcal{S}, \mu)$ is a measure space and $f:X\rightarrow > [0,\infty]$ is an $\mathcal{S}$-measurable function. Prove that $$\int f\,d\mu>0 \text{ if and only if } \mu(\{x\in X: f(x)>0\})>0.$$

For a measure space $(X,\mathcal{S},\mu)$ and a measurable function $f:X\rightarrow[0,\infty]$, my textbook defines $\int fd\mu$ as $$\sup\left\{\sum_{k=1}^m a_k\mu(A_k): \sum_{k=1}^m a_k \chi_{A_k}\leq f, \text{ where } m\in\mathbf{Z}^+, a_1,\ldots, a_m\geq 0 \text{ and } A_1,\ldots, A_m\in\mathcal{S}\right\}.$$ Thus I know for the $\implies$ direction, I have that this $\sup>0$. I see that this $\sup$ is taken over the set of certain sums of measures of subsets of $X$. I just don't know how to draw a connection with this and the condition I want to prove, that being $\mu(\{x\in X: f(x)>0\})>0.$ For the $\impliedby$ direction, I am just as lost, but I think understanding the first direction could make the converse easier for me to understand.

So, could any of you provide me with a hint, or maybe help me understand in a more intuitive way what this question is asking? I would really appreciate it!

Edit: I do believe it is assumed that $f$ is a nonnegative function.

Answer 1

Hint

Assume that $0\le a_k\chi_{A_k}\le f.$ Then we have that

$$a_k\mu(A_k)\le a_k\mu(\{x\in X: f(x)>0\}).$$

What do you get if $a_k>0$ and $\mu(\{x\in X: f(x)>0\})\le 0?$

Answer 2

By Tonelli's theorem, $$ \int_X f d\mu = \int_X\int_0^f d\lambda \ d\mu =\iint_{X\times[0,\infty]} \chi_{\{x,t: f(x)>t\}} d(\mu\otimes \lambda) = \int_0^\infty \mu(f>t) \ d\lambda(t),$$ where $\lambda$ is Lebesgue measure on $(0,\infty)$. if $LHS>0$ then $RHS>0$ so that there is at least one $t$ for which $\mu(f>t) > 0$. This is decreasing in $t$ since the sets $\{f>t\} = f^{-1}((t,\infty))$ and $(t,\infty)$ are nested. Thus $\mu(f>0) \ge \mu (f>t)>0$. A similar(but not exactly the same!) idea also works for the converse.

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Judging by the content of your post, you don't yet know the proof of Tonelli's theorem to justify the exchange of integrals above...but IMO this integral identity gives good intuition.