Prove that lim $\int_{E_n}f d\mu = 0$

Question

Well they give me the following statement:

Be $f$ integrable in $(X,F,\mu)$ and {$E_n$} $\subset F$ as {$E_n$} $ \downarrow E$ with $\mu(E) = 0$. Prove that $\lim_{n \to \infty} \int_{E_n} f d\mu = 0$

Well my idea to prove this is use the DCT (Dominated Convergence Theorem) as $f*\mathbb{1_{E_n}} \rightarrow f*\mathbb{1_{E}}$ and |$f*\mathbb{1_{E_n}}$| $< f$ I use the DCT and:

$\lim \int_{E_n} f d\mu = \lim \int f*\mathbb{1_{E_n}} d\mu = \int_{X} \lim f*\mathbb{1_{E_n}} d\mu = \int_{X} f*\mathbb{1_{E}} d\mu = f*\mu(E) = 0$

I don't know if this is correct or not.

Answer 1

Let $(X,\mathcal F, \mu)$ be a measure space and let $f$ be nonnegative and integrable. For every $E \in \mathcal F$, define a new measure $\nu$ by $$\nu(E) = \int_E f\,d\mu$$ In your case, since $(E_n) \downarrow E$, so $E_{n+1} \subset E_n$, under the hypothesis that one of the $E_n$ has finite measure, thanks to the continuity from above of $\nu$ we have

$$ \lim_{n\to \infty} \int_{E_n} f d\mu = \lim_{n\to \infty} \nu(E_n) = \nu\left(\bigcap_nE_n \right) = \nu(E) = \int_E f d\mu = 0$$

Answer 2

Let $\lambda A = \int_A |f| d \mu$. Then $\lambda$ is a finite measure since $f$ is integrable. In particular, it is continuous from above, so $ \lambda E_n \downarrow \lambda E = 0$. Since $\lambda E_n \ge |\int_{E_n} f d \mu| \ge 0$ we see that $\int_{E_n} f d \mu \to 0$.