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Let $(X,M,\mu)$ be a measure and let $\{f_n\}$ be a sequence that is both uniformly integrable and tight on $X$. Suppose $f_n\to f$ a.e. and $f$ is finite a.e. How can we show that $f$ is integrable on $X$.

My approach is to write $X$ as $E_1\cup(X\setminus E_0)\cup (E_0\setminus E_1) $ . Then apply Fatou's lemma to conclude show that $f$ is integrable on $X\setminus E_1$. Since I have that $\mu(E_0)\lt \infty$ I can apply Fatou's lemma and Egorov, to show that $f$ is integrable on $E_1$ and $E_0\setminus E_1$.

Does this seem right?

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up vote 2 down vote accepted

I'm not sure what your sets are, but it seems you have the right idea. Here is an outline of one proof:

First prove the

Lemma: If $f_n\rightarrow f$ and $\int_E|f_n|\le M$ for each $n$, then $\int_E|f|\le M$. (Use Fatou.)

Having done that, proceed to prove the main result as follows:

Step 1) Use tightness to find a set $A$ of finite measure such that $\int_{A^C} |f_n|\le1$ for each $n$. Note by the Lemma then, that $\int_{A^C} |f|\le 1$.

Step 2) Use uniform integrability to find a $\delta>0$ such that whenever $\mu(E)<\delta$ then $\int_E |f_n|\le 1$ for each $n$. By the Lemma, then, we would have $\int_E |f|\le1$ for such $E$.

Step 3) Apply Egoroff to find $B\subset A$ with $\mu(A\setminus B)<\delta$ such that $f_n$ converges uniformly on $B$. Now pick $N$, so that for $m,n\ge N$, $\int_B|f_n-f_m|\le1$.

Step 4) From step 3) we have $\int_{ B}|f_n|\le 1+\int_X|f_N|$ for each $n\ge N$. Thus, there exists an $M$ so that $\int_{ B}|f_n|\le M$ for all $n$. Use the Lemma once more to conclude that $\int_{ B}|f|\le M$.

Step 5) Conclude that $\int_X |f| =\int_{A^C}|f|+\int_{A\setminus B}|f|+\int_B|f|<\infty$.

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Thank you very much. What other ways exist of proving it? – Jack Mar 27 '12 at 23:28
@Jack This is the only method that I'm aware of. But, try Googling "Vitali Convergence Theorem". – David Mitra Mar 27 '12 at 23:31
Thanks. I'll do that. – Jack Mar 27 '12 at 23:35
One more thing. How is the finiteness of $f$ used in the poof? – Jack Mar 27 '12 at 23:47
@Jack It's needed for Egorov's Theorem. – David Mitra Mar 27 '12 at 23:58

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