# Uniform integrability and tightness.

Definition: Let $(X,M,\mu)$ be a measure space and $\{f_n\}$ a sequence of measurable functions on $x$ that are integrable.
Then $\{f_n\}$ is uniformly integrable if for every $\epsilon >0$, there is a $\delta >0$ such that if $E$ is a measurable subset of $X$ such that $\mu(E) < \delta$, then $$\int_E |f_n|~d\mu < \epsilon\qquad\text{for every} ~n.$$

$\{f_n\}$ is said to be tight if for each $\epsilon >0$, there is a subset $X_0$ of $X$ such that $\mu(X_0)< \infty$ and $$\int_{X\setminus X_0} |f_n|~d\mu < \epsilon\qquad\text{for every} ~n.$$

Theorem:(Vitali Convergence) Let $(X,M,\mu)$ be a measure space. Let $\{f_n\}$ be a sequence of uniformly integrable functions that also forms a tight sequence. Suppose $f_n(x) \to f(x)$ a.e. on $X$. Then, $f$ is integrable and, $$\lim_{n\to \infty} \int_X f_n~d\mu = \int_X f~ d\mu.$$

I wish to prove the following:

Let $\{f_n\}$ be a sequence of non-negative integrable functions on $X$. Suppose that $\{f_n(x)\} \to 0$ for almost all $x\in X.$. Then $$\lim_{n\to\infty} \int f_n~d\mu =0 \Leftrightarrow \{f_n\}~\text{is uniformly integrable and tight.}$$

This is my Attempt:

$(\Leftarrow)$ Suppose $f_n \to 0$. If $\{f_n\}$ is uniformly integrable and tight, then by Vitali's Convergence theorem, $\lim_{n\to \infty} \int f_n~d\mu = 0$.

$(\Rightarrow)$ Let $\lim_{n\to \infty} \int f_n~d\mu = 0$. Let $\epsilon >0$. Then $\exists$ an $N$ such that $\int_X f_n ~d\mu< \epsilon$ whenever $n\geq N.$ Also, since $f_n \geq 0$, if $E$ is a measurable subset of $X$ and $n\geq N$, then $\int _E f_n~d\mu < \epsilon.$

I know that if I have a finite sequence $\{f_k\}_{n=1}^N$ of non-negative integrable functions over $X$, then $\{f_k\}_{n=1}^N$ is uniformly integrable, since if $E\subset X$ and $\mu(E)<\delta_k>0$ then $\int_E f_k~d\mu < \epsilon$. I can take $\delta=\min (\delta_1,\ldots, \delta_k)$ so that $\mu(E)< \delta$ and $$\int_E f_k~d\mu < \epsilon.$$

I'm afraid this is where I'm stuck and I don't know how to proceed. Any form of help will be very much appreciated. Thanks.

In order to show tightness, fix $\varepsilon>0$. Then you get $N=N(\varepsilon)$ such that $\int_Xf_nd\mu\leq\varepsilon$ if $n\geq N+1$. Now, for all $n\leq N$, you can find a positive $M$ such that $\int_{\{f_n\geq M_n\}}f_nd\mu\leq \varepsilon$, using integrability of $f_n$. (if $f$ is integrable apply the monotone convergence theorem to $f\mathbf 1_{\{|f|\geq n\}})$

Put $A_n:=\{f_n\leq M_n\}$, then $A_n$ is measurable. Take $X_0:=\bigcap_{k=1}^NA_k^c$. Each $A_k^c$ has finite measure (since $\mu(A_k^c)\leq \frac 1{M_k}\int f_kd\mu$) so $X_0$ is of finite measure. Check that we have the wanted inequality.

– Kuku
Mar 5 '12 at 23:56
• It seems correct. Mar 6 '12 at 9:49
• Thanks. I have a couple of questions regarding tightness: (1) can u please explain your second statment? and (2) how do we know that $A_k^c$ has finite measure?
– Kuku
Mar 7 '12 at 11:41
• $A_k^c$ has a finite measure, otherwise $f_k$ wouldn't be integrable. If $f$ is a non-negative integrable function, then the sequence $\{f\mathbf 1_{f\leq n}\}$ increases to $f$ and by the monotone convergence theorem the integral of $f_n$ converges to the integral of $f$. Mar 7 '12 at 11:52
• Is this how we get the inequality: $\int_{X\setminus X_0} f_n~d\mu < \int_X f_n ~d\mu < \epsilon$?
– Kuku
Mar 7 '12 at 14:22

When you're trying to prove $(\Rightarrow)$, the limit gives you a way to bound the integrals of $f_n$ by $\epsilon$ for sufficiently large $n$. Then it's a matter of using the fact that any collection of finite number of $L^1(E)$ functions are both uniformly integrable and tight.

This is problem 1 in $\textit{Royden and Fitzpatrick}$ page 99. In the errata the author mentions to interchange problem 1 and 2 because problem 2 states to prove that any collection of finite number of $L^1(E)$ functions are both uniformly integrable and tight over $E$. Once you prove that the problem becomes trivial.

This solution is for readers of Real Analysis, fourth edition, by Royden and Fitzpatrick.

This is analogous to Theorem 24 of Section 4.6. If $$\{h_n\}$$ is uniformly integrable and tight, then $$\lim_{n\to\infty}\int_E h_n = 0$$ by the Vitali convergence theorem. Conversely, assume $$\lim_{n\to\infty}\int_E h_n = 0$$. Then $$\{h_n\}$$ is uniformly integrable over $$E$$ by Theorem 26 of Section 4.6 noting that the converse part of the proof does not require that $$E$$ be of finite measure. We now show tightness. For each $$\epsilon > 0$$, we may choose an index $$N$$ for which $$\int_E h_n <\epsilon$$ if $$n\ge N$$. Therefore, because $$h_n\ge 0$$ on $$E$$,

if $$E_0$$ is a subset of $$E$$ of finite measure and $$n\ge N$$, then $$\displaystyle\int_{E\sim E_0} h_n <\epsilon.\quad$$(*)

According to Problem 1, the finite collection $$\{h_n\}_{n=1}^{N-1}$$ is tight over $$E$$. Let $$E_0$$ respond to the $$\epsilon$$ challenge regarding the criterion for the tightness of $$\{h_n\}_{n=1}^{N-1}$$. We infer from (*) that $$E_0$$ also responds to the $$\epsilon$$ challenge regarding the criterion for the tightness of $$\{h_n\}$$.