# measure zero

• Dec 6th 2009, 02:07 PM
poincare4223
measure zero
Let $(X, \mathcal{B}, \mu)$ be a measure space and $f : X \rightarrow \mathbb{R}\cup \{ \infty \}$ be a non-negative integrable function. Prove that $\mu (\{ x : f(x)=\infty \}) =0$.

This seems pretty simple. However, I don't know how to prove that a set has measure zero. I know that if it is integrable then the integral is finite. However, I don't see how the set would have to have measure zero right now.
• Dec 9th 2009, 03:14 AM
Moo
Hello,

Let $A_n=\{x\in X ~:~ f(x) \geq n\}$

We have $f_n=f \cdot \bold{1}_{A_n} \xrightarrow[n\to\infty]{} 0$ and we also have $|f_n|\leq f$, which is integrable.
So by Lebesgue's dominated convergence theorem, we have $\int_{A_n} f ~d\mu=\int_X f_n ~d\mu \xrightarrow[n\to\infty]{} 0$

Then note that $\int_X f \cdot\bold{1}_{A_n} ~d\mu=\int_X f \cdot \bold{1}_{f\geq n} ~dmu\geq \int_X n \cdot \bold{1}_{A_n} ~d\mu=n\mu(A_n)>0$

Since the first term goes to 0, and that the last term is >0, we get that $\lim_{n\to\infty} n \mu(A_n)=0$

So obviously, $\lim_{n\to\infty} \mu(A_n)=0$

And finally, see that $\{f=\infty\}=\bigcap_{n\geq 1} A_n$

and you should be done...