Question about Lebesgue integral When I prove a real analysis problem, I need a theorem about Lebesgue integral, but I cannot find this theorem in any standard reference, intuitively, I think it is correct, but I do not know how to prove it, can you give me some hint or construct a counterexample?
$(X, \Omega,\mu)$ is a measure space and $\Omega=\sigma(A)$, which means $\Omega$ is the smallest $\sigma$ algebra generated by $A$. $f$ is Lebesgue integral.
If for every $B \in A$, we have $\int_B f=0$, then for every $B\in \Omega$, we also have $\int_B f=0$

Edit The example I gave this morning was fun but very stupid; there are much more trivial examples. Let $X=\{0,1\}$, say every subset of $X$ is measurable, define $\mu(\{0\})=0$ and $\mu(\{1\})=1$, let $A=\{\{0\}\}$
and let $f=1$. Then $\int_Ef\,d\mu=0$ for every $E\in A$, although $\int_X f\,d\mu=1$.
The original is a $\sigma$-algebra that gives examples of various things, although it's not needed here:
Original Let $X$ be any uncountable set. Let $\Omega$ be the class of all $E\subset X$ such that either $E$ or $X\setminus E$ is countable (you should verify that $\Omega$ is in fact a $\sigma$-algebra). Let $A$ be the class of countable subsets of $X$; then it's clear that $\Omega=\sigma(A)$.
Define $\mu(E)=0$ if $E$ is countable, $\mu(E)=1$ if $X\setminus E$ is countable. Verify that $\mu$ is a measure.
Now let $f=1$. Then $\int_Ef\,d\mu=0$ for every $E\in A$, although $\int_X f\,d\mu=1$.
