Nonnegative measurable function on measure zero set Let $f$ be a nonnegative measurable function from $X$ to $R$.
Then, for $\mu$$(A)=0$, $\int_A f d\mu$=0?
I think by using increasing simple function to $f$ and Monotone convergence theorem,
it's true but i'm not sure...
 A: In this case, it would be true by definition, as you're only interested in simple functions (think what is $\int_A f d\mu$). 
A: Let $s=\sum_{i=1}^na_i\chi_{A_i}$ be a simple function with $0\leq s\leq f$. Then, $$\int_As\,d\mu=\sum_{i=1}^na_i\mu(A\cap A_i)=0.$$ Since this holds for all those $s$, you have that $\displaystyle\int_Af\,d\mu=0$.
A: Let ${s_n}$ be a sequence of simple functions such that $s_{n+1}\geq s_n \geq 0$ for all $n\in \mathbb N$ and $s_n \to f$ a.e. Via the Monotone convergence theorem, we have $$\int_Afd\mu=\lim\limits_{n\to \infty}\int_As_nd\mu. \ \ \  \ \ \ (1)$$Since $s_n$  is a simple function we infer $s_n=\sum\limits_{k=1}^{m_n}a_{n_k}1_{A_{nk}}$ where the sets $A_{nk}$ are measurables and $1_{A_{nk}}$ is the characteristic function of $A_{nk}$, in consequence $$\int_As_nd\mu=\sum\limits_{k=1}^{m_n}a_{n_k}\mu(A_{nk}\cap A)=0, \ \ \ (2)$$ because $0\leq\mu(A_{nk}\cap A)\leq \mu(A)=0$. From (1) and (2) we conclude $$\int_Afd\mu=0$$
A: I could be missing something obvious, but...
$f1_A = f1_{\emptyset} \ \text{P-a.s.}$ and $f1_{\emptyset} = 0$
$\implies E[f1_A] = E[f1_{\emptyset}] = E[0] = 0$
Edit: Ah I see what I'm missing. The 2 statements are actually equivalent. Yeah just refer to the other answers.
'Almost sure equality implies same expectation' is equivalent to 'integral over null set is zero.' What's equivalent to 'almost sure implies (...)'?
Integral over null set is zero
