# measure theory exercise: null integral implies null function

Let $\Omega \subset R^n$ a non empty open set and $f: \Omega \rightarrow R$ a nonnegative measurable function with $\int_{\Omega} f =0$. Then $f=0$ in $\Omega$ almost everywhere.

I have no idea of how to start this problem, someone could help me ?

My try (I am not sure):

Let $E_n:= \{ x \in \Omega; f(x) > 1 / n\}, n \in N$ and define $E:= \{ x \in \Omega; f(x) > 0\} = \cup_{n \geq 1} E_n.$

Note that

$$0 = \int_{\Omega} f \geq \int_{E} f \geq \int_{E_n} f \geq \frac{|E_n|}{n} \geq 0.$$

Then $|E_n| = 0$ for all n, which implies $|E| = 0.$ Then $f=0$ in $\Omega$ a.e

I am not sure because it seems that we can replace the set $\Omega$ by a measurable set with zero measure and if we consider a set like this the affirmation is not true.

• it seems missing something, example, we may suppose that $\Omega$ has not null measure and $f$ is bounded. This makes the exercise obvious, but, I don't know, is missing anything? – L.F. Cavenaghi Feb 25 '16 at 23:02
• @GaussTheBauss thanks, read it to fast – Lionel Ricci Feb 25 '16 at 23:04
• @LionelRicci nonnegative – GaussTheBauss Feb 25 '16 at 23:05
• the sine function is negative in $(\pi , 3\pi /2)$. This function does not satisfy the hipothesis – math student Feb 25 '16 at 23:06
• probably you mean f=0 almost everywhere, right? – anonymous Feb 25 '16 at 23:07

Hint: Consider the sets $\{ x\in \Omega : f(x) > 1/k\}, k = 1,2,\dots$
Suppose that $f>0$ on some subset $S\subset \Omega$ of strictly positive measure. Then $$0<\int_{S}f \leq \int_{\Omega}f = 0\ .$$ This contradiction shows that $f = 0$ on every subset of strictly positive measure.