
measure zero
Let $\displaystyle (X, \mathcal{B}, \mu)$ be a measure space and $\displaystyle f : X \rightarrow \mathbb{R}\cup \{ \infty \}$ be a nonnegative integrable function. Prove that $\displaystyle \mu (\{ x : f(x)=\infty \}) =0$.
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.

Hello,
Let $\displaystyle A_n=\{x\in X ~:~ f(x) \geq n\}$
We have $\displaystyle f_n=f \cdot \bold{1}_{A_n} \xrightarrow[n\to\infty]{} 0$ and we also have $\displaystyle f_n\leq f$, which is integrable.
So by Lebesgue's dominated convergence theorem, we have $\displaystyle \int_{A_n} f ~d\mu=\int_X f_n ~d\mu \xrightarrow[n\to\infty]{} 0$
Then note that $\displaystyle \int_X f \cdot\bold{1}_{A_n} ~d\mu=\int_X f \cdot \bold{1}_{f\geq n} ~dmu\geq \int_X n \cdot \bold{1}_{A_n} ~d\mu=n\mu(A_n)>0$
Since the first term goes to 0, and that the last term is >0, we get that $\displaystyle \lim_{n\to\infty} n \mu(A_n)=0$
So obviously, $\displaystyle \lim_{n\to\infty} \mu(A_n)=0$
And finally, see that $\displaystyle \{f=\infty\}=\bigcap_{n\geq 1} A_n$
and you should be done...