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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{equation} \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} \end{equation}

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?

Thanks, in advance.

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3 Answers 3

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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 $\mu(|f_n-f|) \to 0$.

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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.

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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.

To answer your questions:

(1) Yes, your approach is correct.

(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.

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