TITLE: $\int_A fd\mu=0$ for all $A$ in a generator of the $\sigma$-algebra $\Rightarrow$ $f=0$ $\mu$-almost everywhere?
QUESTION [6 upvotes]: Assume I have a measurable signed function $f$ for which the integral with respect to a measure $\mu$, on some measurable sets $\mathcal{C}$ generating the sigma algebra $\mathcal{E}$, is zero (maybe a $\pi$-system). Is this enuough to imply the function is zero almost everywhere, if not what is needed?
Thanks.

REPLY [3 votes]: No, not in general. Let $\Omega$ any uncountable set and
$$ \mathcal E = \{A\subset \Omega: \text{$A$ or $A^c$ is countable}\}=\sigma(\mathcal C)$$
with
$$ \mathcal C:=\{\{\omega\}:\omega\in\Omega\}\cup \{\emptyset\}.$$
It is easy to check that
$$ \mu\colon\mathcal E\to [0,1], \quad A\mapsto \begin{cases}0, & \text{$A$ countable,} \\ 1, & \text{$A^c$ countable,}\end{cases}$$
is a measure on $\mathcal E$. Any measurable $f\colon\Omega\to\mathbb R$ fulfills
$$ \int_Afd\mu=0 \quad \text{for all $A\in\mathcal C$.}$$
Note that $\mathcal C$ is even closed under intersections.

However, if $\mu$ is $\sigma$-finite, $f$ is non-negative (or non-positive), $\mathcal C$ is stable under intersections and contains a sequence $(C_n)_{n\ge1}$ of sets such that $C_n \uparrow \Omega$ and $\mu(C_n)<\infty$, the statement follows from the uniqueness theorem for measures.