$\int_A fd\mu=0$ for all $A$ in a generator of the $\sigma$-algebra $\Rightarrow$ $f=0$ $\mu$-almost everywhere?

Question

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.

Answer 1

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.

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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.

Answer 2

I think that you the answer to the question is positive, provided you add the assumption that $\int_\Omega f \mathrm{d}\mu=0$.

In other words, suppose $\mathcal{C}$ is a $\pi$-system of generators for the sigma-algebra $\mathcal{E}$. Suppose $\int_A f \mathrm{d}\mu=0$ for all $A\in \mathcal{C}$ and for $A=\Omega$. Then, $f=0$ $\mu$-a.e.

Indeed, the set $\Lambda=\{A\in \mathcal{E}\colon \int_A f\mathrm{d}\mu=0\}$ is a $\lambda$-system:

• it contains $\Omega$ by assumption;

• it is closed under complements: if $A\in\Lambda$, then $\int_{A^c}f \mathrm{d}\mu = \int_{\Omega}f \mathrm{d}\mu - \int_{A}f \mathrm{d}\mu =0-0=0$ and $A^c\in\Lambda$;

• it is closed under unions of pairwise disjoint sets: if $A_1,A_2, \ldots\in \Lambda$ are pairwise disjoint, then $\int_{\cup_n A_n} f \mathrm{d}\mu = \sum_{n} \int_{A_n} f \mathrm{d}\mu =0$ and $\cup_n A_n\in\Lambda$.

As $\mathcal{C}$ is a $\pi$-system and $\Lambda$ is a $\lambda$-system such that $\mathcal{C}\subseteq\Lambda$ (by assumption), we have $\mathcal{E}=\sigma(\mathcal{C})\subseteq\Lambda$ by Dynkin's $\pi$-$\lambda$ theorem. This means that $\int_A f \mathrm{d}\mu =0$ for all $A\in\mathcal{E}$. Taking $A^+:=\{f\geq 0\}\in\mathcal{E}$ and $A^-:=\{f\leq 0\}\in\mathcal{E}$, we obtain that $f^+$ and $f^-$ are both zero $\mu$-a.e., so $f$ also is.