# measure theory question

• Oct 15th 2008, 06:25 AM
rogerpodger
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
• Oct 15th 2008, 11:11 AM
Moo
Hello,

Here is a proof..

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

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

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

__________________________________________
$\underline \Rightarrow \quad :$

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

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

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

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

$\underline \Leftarrow \quad :$

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

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

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

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

(Whew)
• Oct 15th 2008, 11:51 AM
Laurent
Here's another proof for the more difficult implication (a more direct one, without simple functions):

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