Lebesgue measure problem

Question

Let $f$ be a non-negative measurable function on $\mathbb{R}$, and suppose that $\int f=0$. Prove that the set where $f \neq 0$ is a zero set.

The hint says to let $E_n=\{f>1/n\}$ and then compare $U(f)$ with $U(f|_{E_n})$.

I need help to prove this and compare the two above.

Answer 1

Let $E_n=\{f\ge 1/n\}$ and $\chi_{E_n}$ be the indicator function of this set. Then clearly $f\ge \chi_{E_n}f\ge 0$, so $$\int f\ge \int\chi_{E_n}f\ge 0, $$ but $$0\ge\int \chi_{E_n}f\ge \frac{|E_n|}{n}\ge 0.$$

Can you conclude now?

Answer 2

You can argue like this:

$0 \leq \frac{1}{n} \mu \left( E_{n}\right) = \int_{E_{n}} \frac{1}{n} \leq \int _{E_{n}} f = 0$

Hence $\mu \left( E_{n}\right)=0$ for any $n$. Then the set you are considering is $\cup_{n} E_{n}$ and by continuity of measure (the union is growing, since $E_{n} \subset E_{n+1}$) you can conclude the measure of your set is $0$.

Answer 3

Probably the most straightfoward way to do this is by Chebyshev's inequality (and it helps to know Chebyshev anyway):

Try proving that for $\lambda >0$, $\displaystyle m(\{x : f(x)\geq \lambda\}) \leq \frac{1}{\lambda} \int f$. (of course, for $f\geq 0$)

Then note that $E= \{x \in \mathbb{R} : f(x)>0\} = \bigcup_{n=1}^\infty E_n$ where $E_n = \{x : f(x)\geq \frac{1}{n}\}$. Now, apply Chebyshev to conclude that $m(E) = 0$.