Proof of Vitali's Convergence Theorem

Question

This is an exercise from Rudin's Real and Complex Analysis.

Prove the following convergence theorem of Vitali:

Let $\mu(X)\lt \infty$ and suppose a sequence of functions, $\{f_n\}$ is uniformly integrable, $f_n(x)\to f(x)$ a.e. as $n\to \infty$, then $f\in L^1(\mu)$ and $$ \lim_{n\to\infty} \int_X |f_n-f|~d\mu = 0.$$

Attempt:

Since $f_n$ is uniformly integrable, $\exists~\delta \gt 0$ such that whenever $\mu(E)\lt \delta$, we have $$\int_E |f_n|~d\mu \lt \frac{\varepsilon}{3} \quad \forall~n.$$ Since $\mu(X)\lt \infty$, Egoroff says that we can find a set $E$ such that $f_n \to f$ uniformly on $E^c$ and $\mu(E)\lt \delta$. So $\exists$ an $N$ such that for $n\gt N$ $$\int_{E^c} |f_n-f|~d\mu\lt \frac{\varepsilon}{3}.$$ So, \begin{align*} \int_X |f_n - f| ~ d\mu & = \int_{E^c} |f_n - f| ~ d\mu + \int_E |f_n - f| ~ d\mu \\ & \leq \int_{E^c} |f_n - f| ~ d\mu + \int_E |f| ~ d\mu + \int_E |f_n| ~ d\mu \\ & < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ & = \varepsilon. \end{align*}

Now to show that $f\in L^1(\mu)$, I have to show that $\int_X |f|\lt \infty.$ Somehow I feel I have to use Egoroff again but I'm kind of lost. I'd be grateful if someone could look over what I've done above and see if it's okay and perhaps provide a little help with showing the $f\in L^1(\mu)$.

Thanks.

What you have so far implies $f \in L^1(\mu)$, since $|f| \le |f_n| + |f_n - f|$ and you know that the latter are integrable. However, there's a small gap in what you have: why is $\int_E |f|,d\mu < \epsilon/3$? — Nate Eldredge, Feb 24 '12 at 21:22
Also, I think in your second sentence, $\mu(E^c) < \delta$ should be $\mu(E) < \delta$. — Nate Eldredge, Feb 24 '12 at 21:24
@NateEldredge, (To you first comment) That can be handled with Fatou's Lemma. — leo, Feb 24 '12 at 21:33
@NateEldredge: I thought that followed from uniform integrability... — Kuku, Feb 24 '12 at 21:44
@kuku, I think you can argue as you did. Then, you can prove that $f\in L^1$ as Nate suggest. See my answer — leo, Feb 24 '12 at 22:47

leo · Accepted Answer · 2018-01-01T18:43:00.183

16

Your first two steps are fine to me (Use of uniform integrability and Egoroff's Theorem).

Note that in general if $f_n\to f$ and $\int_E |f_n|\leq M$ for some $M$, by Fatou's Lemma you have $$ \int_E |f|= \int_E \liminf |f_n| \leq \liminf \int_E |f_n|\leq M. $$

To finish your proof you must say:

So, for any $n\geq N$ (the $N$ in your post) $$\begin{align*} \int_X |f_n-f|~d\mu & = \int_{E} |f_n-f|~d\mu +\int_{E^c} |f_n-f|~d\mu\\ & \leq \int_{E} |f_n-f|~d\mu + \int_{E^c} |f|~d\mu + \int_{E^c} |f_n|~d\mu\\ & \lt \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3}\\ & =\varepsilon. \end{align*}$$ The second step is justified by the triangle inequality and the observation made at the beginning of this post.

To justify that $f\in L^1$, see Nate's comment.

edited Jan 01 '18 at 18:43

answered Feb 24 '12 at 21:56

leo

10,769

i guess, you accidentally interchanged $E$ and $E^c$ in the second step. – derivative Nov 28 '13 at 14:58
@derivative No. Why you say so? It looks okay to me. – leo Nov 28 '13 at 21:36
in the post of Kuku is $\int_{E} |f_n|~d\mu<\epsilon/3$
in your post is $\int_{E^c} |f_n|~d\mu<\epsilon/3$
– derivative Nov 28 '13 at 21:46
1

If you look closely the roles of $E$ and $E^c$ are reversed in my answer, respect to those in the kuku's post. – leo Nov 28 '13 at 21:53
ok after your comment about fatou it makes sense. but why did you change them ? – derivative Nov 28 '13 at 21:58
@derivative By that time I was used to write stuff like this that way – leo Nov 28 '13 at 22:24
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But where is the condition $|f|<\infty$ used? – mnmn1993 Mar 29 '18 at 15:09
@mnmn1993 To show that $\int_{E^c}|f|d\mu<\varepsilon/3$, $f$ finite almost everywhere implies $f\chi_{E^c}$ is in $L^1$, and so $\delta$ can be taken small enough so that that integral is controlled also, see e.g. Problem 1.12 in Rudin RCA. – Eli Seamans Dec 12 '24 at 02:58

Proof of Vitali's Convergence Theorem

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