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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.

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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.

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    $\begingroup$ thanks for your answer. what about the uniform integrability part? $\endgroup$
    – Kuku
    Mar 5 '12 at 23:56
  • $\begingroup$ It seems correct. $\endgroup$ Mar 6 '12 at 9:49
  • $\begingroup$ 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? $\endgroup$
    – Kuku
    Mar 7 '12 at 11:41
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    $\begingroup$ $A_k^c$ 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 $f$. $\endgroup$ Mar 7 '12 at 11:52
  • $\begingroup$ Is this how we get the inequality: $\int_{X\setminus X_0} f_n~d\mu < \int_X f_n ~d\mu < \epsilon$? $\endgroup$
    – Kuku
    Mar 7 '12 at 14:22
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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.

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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 $\{h_n\}$.

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