Nonnegative measurable function on measure zero set

Question

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...

Answer 1

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 $\displaystyle\int_Af\,d\mu=0$.

Answer 2

In this case, it would be true by definition, as you're only interested in simple functions (think what is $\int_A f d\mu$).

Answer 3

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 $$\int_Afd\mu=0$$