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 $n\to \infty$, and $|f(x)|\lt \infty$ a.e., then $f\in L^1(\mu)$ and $$ \lim_{n\to\infty} \int_X |f_n-f|~d\mu = 0.$$


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, $$\begin{align*} \int_X |f_n-f|~d\mu & = \int_{E^c} |f_n-f|~d\mu +\int_E |f_n-f|~d\mu\\ & \leq \int_{E^c} |f_n-f|~d\mu + \int_E |f|~d\mu + \int_E |f_n|~d\mu\\ & \lt \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3}\\ & =\varepsilon. \end{align*}$$

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 $f\in L^1(\mu)$.


  • 3
    $\begingroup$ 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 $\int_E |f|\,d\mu < \epsilon/3$? $\endgroup$ Feb 24 '12 at 21:22
  • $\begingroup$ Also, I think in your second sentence, $\mu(E^c) < \delta$ should be $\mu(E) < \delta$. $\endgroup$ Feb 24 '12 at 21:24
  • 2
    $\begingroup$ @NateEldredge, (To you first comment) That can be handled with Fatou's Lemma. $\endgroup$
    – leo
    Feb 24 '12 at 21:33
  • $\begingroup$ @NateEldredge: I thought that followed from uniform integrability... $\endgroup$
    – Kuku
    Feb 24 '12 at 21:44
  • 2
    $\begingroup$ Prove that $f$ is in $L_1$ first... $\endgroup$ Feb 24 '12 at 21:49

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.

  • $\begingroup$ i guess, you accidentally interchanged $E$ and $E^c$ in the second step. $\endgroup$
    – derivative
    Nov 28 '13 at 14:58
  • $\begingroup$ @derivative No. Why you say so? It looks okay to me. $\endgroup$
    – leo
    Nov 28 '13 at 21:36
  • $\begingroup$ in the post of Kuku is $\int_{E} |f_n|~d\mu<\epsilon/3$ in your post is $\int_{E^c} |f_n|~d\mu<\epsilon/3$ $\endgroup$
    – derivative
    Nov 28 '13 at 21:46
  • 1
    $\begingroup$ If you look closely the roles of $E$ and $E^c$ are reversed in my answer, respect to those in the kuku's post. $\endgroup$
    – leo
    Nov 28 '13 at 21:53
  • 1
    $\begingroup$ But where is the condition $|f|<\infty$ used? $\endgroup$
    – mnmn1993
    Mar 29 '18 at 15:09

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