1. ## measure theory question

So I have a measurable, non-negative function f. I am to show that the integral of f with respect to x is zero if and only if the set of xsthat make f(x) > 0 is a set of measure zero. To show the first half of this implication, I am using a contrapositive argument. My classmate suggested that I use simple functions to approximate the integral below, but I'm not sure how to do this. Any thoughts?

rogerpodger

2. Hello,

Here is a proof..

$\displaystyle \int f ~ d \mu=0 \quad \Leftrightarrow \quad \mu (\{f > 0\})=0$

Let $\displaystyle (\varphi_n)$ be an increasing positive sequence that converges to $\displaystyle f$, and where $\displaystyle \forall n, ~ \varphi_n$ is a simple function.
We can then say that $\displaystyle \int_E \varphi_n ~ d \mu=\sum_{\alpha \in \varphi_n(E)} \alpha ~ \mu (\{\varphi_n=\alpha\})$

Note that $\displaystyle \int_E \varphi_n ~ d \mu=0 \quad \Leftrightarrow \quad \mu (\{\varphi_n \neq 0\})=0$ (this is very simple to prove with the above formula of $\displaystyle \int_E \varphi_n ~ d \mu$)
Also note that saying $\displaystyle \neq 0$ is equivalent to saying $\displaystyle >0$ since we work on positive functions.

__________________________________________
$\displaystyle \underline \Rightarrow \quad :$

$\displaystyle \int f ~d \mu =0 \Rightarrow \forall n, \int \varphi_n ~d \mu=0$ (because $\displaystyle 0 \leqslant \varphi_n \leqslant f$)

Therefore, $\displaystyle \forall n, ~ \mu(\{\varphi_n > 0\})=0$

But $\displaystyle \varphi_n$ is an increasing sequence. Hence $\displaystyle \{\varphi_n > 0 \}$ is an increasing set.
$\displaystyle \underbrace{\mu(\{\varphi_n > 0 \})}_{=0 ~\forall n} \longrightarrow \mu(\{f > 0\})$

$\displaystyle \Longrightarrow \mu(\{f > 0\})=0$

$\displaystyle \underline \Leftarrow \quad :$

$\displaystyle \mu(\{f > 0\})=0$

Since $\displaystyle \{\varphi_n > 0 \} \subseteq \{f > 0\}$ (because remember that $\displaystyle \varphi_n \uparrow f$), we have $\displaystyle \mu (\{\varphi_n > 0 \}) \leqslant \mu (\{f > 0\})=0 \quad \Rightarrow \quad \mu (\{\varphi_n > 0 \})=0$

Therefore $\displaystyle \int \varphi_n ~ d \mu=0 \quad \forall n$

By the monotone convergence theorem, we can conclude :
$\displaystyle \int f ~ d \mu=\lim_n \int \varphi_n ~ d \mu=0$

3. Here's another proof for the more difficult implication (a more direct one, without simple functions):

Suppose $\displaystyle \int f d\mu=0$, where is $\displaystyle f\geq 0$ is measurable.
Assume by contradiction that $\displaystyle \mu(\{x\,|\,f(x)>0\})>0$.
Choose a sequence $\displaystyle (\varepsilon_n)_n$ strictly decreasing to 0. Since $\displaystyle \{x\,|\,f(x)>0\}=\bigcup_n \{x\,|\, f(x)>\varepsilon_n\}$, where the sets in the union form an increasing sequence, you have: $\displaystyle 0<\mu(\{x\,|\,f(x)>0\})=\lim_n \mu(\{x\,|\, f(x)>\varepsilon_n\})$. As a consequence, there exists $\displaystyle n$ such that $\displaystyle \mu(\{x\,|\, f(x)>\varepsilon_n\})>0$, which implies: $\displaystyle \int f d\mu\geq \varepsilon_n\cdot \mu(\{x\,|\,f(x)>\varepsilon_n\})>0$, in contradiction with the initial assumption.