Function is identically zero almost everywhere

Question

Prove that if $\int_E f d\mu = 0$ for some $f \ge 0$, then $f = 0$ almost everywhere.

This is Execrise 1 in Chapter 11 of baby Rudin.

My attempt:

$\int_E f d\mu = 0 \implies$ sup { ${\int_E s d\mu}=0$ where $s$ is simple integrable and $0\leq s\leq f$

$\implies s=0$ or $\mu (E)=0$

Case 1: $s=0$

Since $f= lim_n\to \infty s_n$ and $s$ is a linear combination of $s_n$, $f=0$.

case 2: $\mu(E)=0$

Let $A_n:=$ {$x\in E : f(x)> 1/n$}

$A_1 \subset A_2 \subset A_3.....$

$A= \cup A_n$

$\mu (A)= \mu (\cup A_n)= lim _n\to \infty \mu (A_n)$

$\mu (A)= 0 \implies\mu (\cup A_n)= lim _n\to \infty \mu (A_n)=0$

But I am unable to proceed further. please suggest how to prove $f=0$ for case 2.

Answer 1

You want to show that $f=0$ a.e. on $E$ right? Since you already have $f \geq 0$, just showing that $\mu(\{x\in E: f(x) >0 \}) = 0$ would suffice. Now $$\{x\in E: f(x) >0 \} = \bigcup_{n\in\mathbb{N}} \{x\in E: f(x) \geq \frac{1}{n} \}$$.

What is $\mu ( \{x\in E: f(x) \geq \frac{1}{n} \})$ for all $n$? Use Markov's inequality.

So what does that imply about $\mu(\{x\in E: f(x) >0 \})$?