2
$\begingroup$

Show that Fatou's lemma, the Monotone Convergence Theorem, the Lebesgue Dominated Convergence Theorem, and the Vitali Convergence Theorem remain valid if "pointwise convergence a.e." is replaced by "convergence in measure"

I have done every part except the part about Vitali's theorem. I tried a google search, but I couldn't find very much, so I am hoping the MSE community would be so kind as to critique my proof. For reference, here is the statement of Vitali's theorem with which I am working:

Let $E$ be of finite measure. Suppose the sequence of functions $\{f_n\}$ is uniformly integrable over $E$. If $f_n \to f$ pointwise a.e. on $E$, then $f$ is integrable over $E$ and $$\lim_{n \to \infty} \int_E f_n = f$$.

Here's my proof. First, I will show that $f$ is integrable. Given $\epsilon = 1$, by uniform integrability there exists $\delta > 0$ such that $A \subseteq E$ measurable with $m(A) < \delta$ implies $\int_A |f_n| < 1$ for every $n \in \Bbb{N}$. Given $\delta$, $E$ can be partitioned into measurable sets $E_1,...,E_k$ with $m(E_i) < \delta$. Hence

$$\int_E |f_n| = \sum_{i=1}^k \int_{E_i} |f_n| < k,$$

for every $n \in \Bbb{N}$, and therefore $\liminf \int_{E} |f_n| \le \sup \int_E |f_n| \le k$. By Fatou's lemma for convergence in measure,

$$\int_E |f| \le \liminf \int_{E} |f_n| \le k < \infty.$$

Now we prove the limit part. Again, by Fatou's lemma for convergence in measure, we get $\int_E f \le \liminf \int_E f_n$. Note that there exists a subsequence $\{f_{n_k}\}$ for which

$$\lim_{k \to \infty} f_{n_k} = \limsup f_n.$$

Since $f_{n_k} \to f$ in measure, there exists a subsequence $\{f_{n_{k_m}}\}$ such that $f_{n_{k_m}} \to f$ pointwise a.e. on $E$. Since $\{f_n\}$ is UI, $\{f_{n_{k_m}}\}$ will also be UI. Hence by Vitali's theorem for pointwise covergence

$$\lim_{m \to \infty} \int_E f_{n_{k_m}} = \int_E f .$$

Since $\{\int_E f_{n_{k_m}}\}$ is a subsequence of $\{\int_E f_{n_k}\}$, it must be the case that

$$\lim_{m \to \infty} \int_E f_{n_{k_m}} = \limsup \int_E f_n,$$

and therefore $\limsup \int_E f_n = \int_E f \le \liminf \int_E f_n$ which implies

$$\lim_{n \to \infty} \int_E f_n = \int_E f$$

$\blacksquare$

Does this sound right?

$\endgroup$
1
$\begingroup$

Looks good, here is a shorter argument with the same idea:

The sequence of numbers $\int f_n \rightarrow \int f$ if and only if for each subsequence $\int f_{n_k}$ there exists a further subsequence $\int f_{n_{k_l}} \rightarrow \int f$.

Let $\int f_{n_k}$ be given, $f_{n_k}\rightarrow f$ in measure, there exists a further subsequence $f_{n_{k_l}} \rightarrow f $ a.e. and $\{f_{n_{k_l}}\}$ is UI, apply Vitali we have $\int f_{n_{k_l}} \rightarrow \int f.$

$\endgroup$
  • $\begingroup$ Oh, wow! This is very nice. Thanks! $\endgroup$ – user193319 Apr 11 '18 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.