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

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

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

Thanks.

Answer 1

You 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 the Nate's comment.