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

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


Does this sound right?


1 Answer 1


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 $\int f_{n_{k_l}} \rightarrow \int f.$

  • $\begingroup$ Oh, wow! This is very nice. Thanks! $\endgroup$
    – user193319
    Apr 11, 2018 at 13:04

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