Rudin's proof of Fatou's Lemma 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!
 A: First, note the fact that 

If $\{a_k\}$ and $\{b_k\}$ are two sequences of real number such that
  \begin{align}
a_k\leq b_k
\end{align}
  for all $k = 1, 2, \ldots$, then 
  \begin{align}
\liminf_{k\rightarrow \infty}\ a_k \leq \liminf_{k\rightarrow \infty}\ b_k.
\end{align}

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}
A: It is that $\lim_{k}g_{k}=\liminf_{n}f_{n}$ and we have $\displaystyle\int\lim_{k}g_{k}=\lim_{k}\int g_{k}$ by Monotone Convergence Theorem.
Now $\displaystyle\int g_{k}\leq\int f_{k}$ and taking limit infimum $k\rightarrow\infty$ both sides we have $\liminf_{k}\displaystyle\int g_{k}\leq\liminf_{k}\int f_{k}$. But $\liminf_{k}\displaystyle\int g_{k}=\lim_{k}\int g_{k}$ because the latter limit exists.
