Let $(X, \mathcal{A},\mu)$ be a measure space and $h:X \to \mathbb{R}, h \geq0$ a measurable function.
Define a map $\mu_h:\mathcal{A} \to \mathbb{R}, \ \mu_h(E):= \int_E h \ d\mu$ for $E \in \mathcal{A}$.
How can it be shown that if $\mu(E)=0 \Rightarrow \mu_h(E)=0$?
To prove this implication I tried:
$\int_E h \ d\mu=\int_E h_+ \ d\mu-\int_E h_- \ d\mu$
So $\int_E h_+ \ d\mu=\int h_+1_E \ d\mu=$sup$\int e \ d\mu$, with $e$ is a simple function.
Here I don't know how to continue to show this implication. Or is there another way to prove it?
Best Answer
By definition $\mu_h(E)=\int h\mathbf1_E d\mu=\sup\{\int sd\mu\mid s\text{ is a measurable simple function with } 0\leq s\leq h\mathbf1_E\}$
So it is enough to prove that $\mu(E)=0$ implies that $\int sd\mu=0$ for every measurable simple function $s$ that satisfies $0\leq s\leq h\mathbf1_E$.
Such function $s$ can be written as a finite sum $\sum_{i=1}^n a_i\mathbf1_{A_i}$ where the $a_i$ are nonnegative and the $A_i$ are measurable subsets of $E$.
So we have $\int sd\mu=\sum_{i=1}^n a_i\mu(A_i)=0$ because $0\leq\mu(A_i)\leq\mu(E)=0$ for every $i\in\{1,\dots,n\}$.