# Uniform Integrability and Convergence

I am trying to solve the following problem: A sequence $$\{f_n\}_{n \in \mathbb{N}}$$ is said to be uniformly integrable in $$L(X,d\mu)$$ if $$\lim_{t \to \infty}\sup_{n\ge1}\int_{\{|f|>t\}}|f_n| \, d\mu = 0.$$ Now, assume that $$\mu(X) < \infty$$ and $$f_n \to f$$ almost everywhere. If $$\{f_n\}_{n \in \mathbb{N}}$$ is uniformly integrable show that $$f \in L(X,d\mu)$$ and $$\lim_{n \to \infty}\int_{X}f_n \, d\mu = \int_{X}f \, d\mu.$$

My Intuitive Idea: Since, we have a finite measure space and pointwise convergent functions, we can use Egorov's Theorem to get uniform convergence. This allows us to get a convergence in the integral except for a set of small measure. Then, use uniform integrability to show that the integral over this small measure set vanishes.

My Attempt: Fix $$\epsilon > 0$$, then there exists a set $$E \subset X$$ such that $$m(X/E) < \epsilon$$ and $$f_n$$ converges $$f$$ uniformly on $$E$$. Then, $$$$\begin{split} \lim_{n \to \infty}\int_{X}f_n \, d\mu &= \lim_{n \to \infty}\int_{E}f_n \, d\mu + \lim_{n \to \infty}\int_{X/E}f_n \, d\mu \\ &= \int_{E}\lim_{n \to \infty}f_n \, d\mu + \lim_{n \to \infty}\int_{X/E}f_n \, d\mu \\ &= \int_{E}f \, d\mu + \lim_{n \to \infty}\int_{X/E}f_n \, d\mu \end{split}$$$$

I do not know how to handle $$\lim_{n \to \infty}\int_{X/E}f_n \, d\mu.$$

My Questions:

(1) Am I on the right track?

(2) What is the intuitive meaning of uniform integrability?

(3) Is the result still true for $$\sigma-$$finite measure spaces?

Here's a somewhat more elementary way which establishes a stronger result :

Since $$\mu(X)$$ is finite, it is easy to show that $$\mu(|f_n|)$$ is bounded using uniform integrability, and so $$\mu(|f|) \leq \liminf \mu(|f_n|) \lt \infty$$ by Fatou's lemma, so that $$f \in L(X,\mu)$$.

It can also be shown that the family $$\{|f_n -f|\}$$ is UI.

Then, note that $$\forall k \in \mathbb{N}$$ $$\mu(|f_n - f|) = \mu(|f_n - f|1_{\{|f_n-f| > k\}}) + \mu(|f_n - f|1_{\{|f_n-f| \leq k\}})$$.

$$\forall \epsilon > 0$$, the first term is smaller than $${\epsilon}/{2}$$ for $$k$$ larger than some $$k'$$ by uniform integrability of $$\{|f_n -f|\}$$. For fixed $$k$$, the second term $$\to 0$$ as $$n \to \infty$$ by the dominated convergence theorem. It follows that $$\forall \epsilon >0$$, $$\mu(|f_n-f|) \leq \epsilon$$ for $$n$$ large enough and so $$\mu(|f_n-f|) \to 0$$.

For the limit of $$\int_{X/E}f_n \, d\mu$$ split the integral into the integral over $$|f|>t$$ and the integral over $$|f|\leq t$$. In the first part use uniform integrability. In the second part use DCT.

Your approach and intuition are correct. To complete the proof, you can use the uniform integrability of the sequence $${f_n}_{n \in \mathbb{N}}$$ to show that the limit of the integral over the set $$X/E$$ is 0.

To do this, notice that for any $$t > 0$$,

$$\int_{{|f_n| > t}} |f_n| , d\mu \ge \int_{X/E} |f_n| , d\mu$$

since the set $${|f_n| > t}$$ includes the set $$X/E$$. Therefore,

$$\lim_{t \to \infty} \sup_{n \ge 1} \int_{{|f_n| > t}} |f_n| , d\mu \ge \lim_{t \to \infty} \sup_{n \ge 1} \int_{X/E} |f_n| , d\mu$$

By the definition of uniform integrability, the left-hand side of this inequality is 0. Therefore,

$$\lim_{t \to \infty} \sup_{n \ge 1} \int_{X/E} |f_n| , d\mu = 0$$

This means that for any $$\epsilon > 0$$, there exists a $$t > 0$$ such that for all $$n \ge 1$$,

$$\int_{X/E} |f_n| , d\mu < \epsilon$$

Hence,

$$\lim_{n \to \infty} \int_{X/E} f_n , d\mu = 0$$

and we can conclude that

$$\lim_{n \to \infty} \int_{X} f_n , d\mu = \int_{E} f , d\mu$$

as desired.

(2) The intuitive meaning of uniform integrability is that the sequence of functions $${f_n}_{n \in \mathbb{N}}$$ becomes "small" (in the sense of the integral over the set $${|f| > t}$$) at a uniform rate as $$t$$ becomes large. This property ensures that the sequence of functions does not have "too much mass" in the large values of $$|f|$$.
(3) The result is still true for $$\sigma-$$finite measure spaces, with the same proof. In this case, the measure space $$(X, \mathcal{F}, \mu)$$ is $$\sigma-$$finite if there exists a countable collection of sets $${E_n}{n \in \mathbb{N}}$$ such that $$X = \bigcup{n \in \mathbb{N}} E_n$$ and $$\mu(E_n) < \infty$$ for all $$n \in \mathbb{N}$$. This is a generalization of the condition that the measure of the space is finite.