I have a question about the proof of Fatou's Lemma in Rudin's Real and Complex Analysis, 3rd ed, on page 23. I underlined the part that I don't follow in red below:
To prove the lemma, I would have to replace the underlined part by:
Then $g_k \le f_n, \forall n\ge k$, so that
$$\int_X g_k d\mu\le \int_X f_n d\mu, \quad \forall n\ge k,$$ and hence
$$\int_X g_k d\mu\le \underset{n\ge k}\inf \int_X f_n d\mu.$$
…
Hence (1) follows from (3). (Take the limit on both sides and apply Monotone Convergence on the left.)
My question: What I don't follow is why the author would say "Hence (1) follows from (3)", when (3) is only
$$\int_X g_k d\mu\le \int_X f_k d\mu, \quad(k=1,2,3,\cdots).$$
I'd appreciate it if someone can point out where I missed. Thanks a lot!
Best Answer
First, note the fact that
In our case, we have that \begin{align} a_k:= \int_X g_k\ d\mu \ \ \text{ and } \ \ b_k := \int_X f_k\ d\mu. \end{align} with \begin{align} \int_X g_k\ d\mu \leq \int_X f_k\ d\mu. \end{align} for all $k$. Hence it follows \begin{align} \liminf_{k\rightarrow \infty} \int_X g_k\ d\mu \leq \liminf_{k\rightarrow\infty}\int_X f_k\ d\mu. \end{align} Next, observe that $\{a_k\}$ is actually a monotonically increasing sequence since $0\leq g_1 \leq g_2 \leq \ldots$. Then, in fact, we have that \begin{align} \liminf_{k\rightarrow \infty}\ a_k = \lim_{k\rightarrow \infty} a_k \end{align} meaning \begin{align} \liminf_{k\rightarrow \infty} \int_X g_k\ d\mu = \lim_{k\rightarrow \infty} \int_X g_k\ d\mu. \end{align} Lastly, by the monotone convergence theorem, we can commute the limit and the integration to get \begin{align} \lim_{k\rightarrow \infty} \int_X g_k\ d\mu = \int_X \lim_{k\rightarrow \infty}g_k\ d\mu = \int_X \lim_{k\rightarrow \infty}\inf_{i\geq k} f_i\ d\mu = \int_X \liminf_{k\rightarrow \infty}\ f_k\ d\mu. \end{align}