I’m practicing for my real analysis exam coming up and am specifically looking at problem 3 in section 3A from Sheldon Axler’s Measure, Integration, and Real Analysis. The question says:
Suppose $(X,S,\mu)$ is a measure space and $f:X\to[0,\infty]$ is an $S$-measurable function. Prove that $$\int fd\mu>0 \quad\text{if and only if}\quad \mu(\{x\in X:f(x)>0\})>0.$$
See definitions 3.1, 3.2, and 3.3 ($S$-partition, lower Lebesgue sum, Lebesgue integral of nonnegative function). I have a solution, but I don't know if my solution is correct, so I was wondering if someone could verify. Here it is:
Proof: Let $F=\{x\in X:f(x)>0\}$.
($\implies$) We prove this direction by contrapositive. Suppose $\mu(F)=0$. Let $P=\{A_1,\dots,A_n\}$ be an arbitrary $S$-partition. Then we have that
$$\begin{align*}
L(f;P)&=\sum_{i=1}^n \mu(A_i)\inf_{A_i}f\\
&=\sum_{A_i\not\subset F}\mu(A_i)\inf_{A_i}f\\
&=\sum_{A_i\not\subset F}\mu(A_i)\cdot 0 = 0.
\end{align*}$$
Since $P$ is arbitrary, this shows every Lebesgue lower sum of $f$ is $0$ so that $\int fd\mu=0$.
($\impliedby$) Suppose $\mu(F)>0.$ Note that since $F=f^{-1}((0,\infty])$, $(0,\infty]$ is a Borel set, and $F$ is $S$-measurable, that $F\in S.$ So then $Q=\{F,X\setminus F\}$ is an $S$-partition, and
$$\int fd\mu\geq L(f;Q)=\mu(F)\inf_{F}f+\mu(X\setminus F)\inf_{X\setminus F}f=\mu(F)\inf_{F}f>0. $$
Best Answer
The last bit is not correct. Notice that $\inf_F f$ can be $0$ so $\mu(F)\inf_F f$ can be $0$, it is not always $>0$ as you said.
The other direction looks good to me though!
Hint: for the case where $\mu(F)>0$ is given you need to use continuity of measure. It will be true that $\mu(f>\epsilon)>0$ for some $\epsilon>0$ and then you can use your same proof strategy, since $\inf f\ge \epsilon>0$ will be true.
"Another" (it is basically the same, just perhaps more conceptual) proof for the first direction might be: $$\int_Xf\,\mathrm{d}\mu=\int_{X\setminus F}f\,\mathrm{d}\mu=\int_{f=0}f\,\mathrm{d}\mu=0$$Where the first equality is valid by virtue of $x\in X\setminus F$ for almost every $x\in X$.