Let $(\Omega,\Sigma,\mu)$ be a measure space and take a strictly positive summable function $f$. For simplicity in what follows let $\mu$ be finite. Now, I am looking for this result (or something similar if this is not true):
For any $A \in \Sigma$ such that $$\int_Af(x)d\mu(x) = 0,$$then $\mu(A) = 0$.
My work would work if the measure was uncountably additive but as we know, measures are $\sigma-$additive (countably additive). The idea was to use Markov's theorem which relates the measure of a level set with the integral of $f$ as for each $a \in (0,\infty)$ we have $$\mu(f > a) \leq \frac{1}{a}\int_Af(x)d\mu(x).$$Now we could write $$\Omega = \bigcup_{a \in (0,\infty)}\{f > a\},$$so that by Markov's theorem $$\mu(A) \leq \mu(\Omega) = \mu\left(\bigcup_{a \in (0,\infty)}\{f > a\}\right) = \sum_{a \in (0,\infty)}\mu(f > a) \leq \frac{1}{a}\int_Af(x)d\mu(x) = 0.$$The result would follow because $\mu$ is assumed to be a positive measure.
This might not even be true so that a counter example is welcome! I am looking for suggestions on this.
Thanks in advance.
We know that a nonnegative function $f$ on $X$ has $\int_X f = 0$ iff $f = 0$. We follow @Andrew's suggestion.
We have $\int_Af(x)\mu(dx) = \int_Xf(x)\chi_A\mu(dx) = 0$ if and only if $f\chi_A = 0$ a.e. But $f > 0$ so $\chi_A = 0$ a.e. this means that $\mu(A) = 0$.