Understanding the proof of the Dominated Convergence Theorem I was going through the proof of the Dominated Convergence Theorem.
Now if we have that $(f_n)$ is a sequence of measurable functions such that $\lvert f_n\rvert\le g$ for all $n$ where $g$ is integrable on $\Bbb R$. And if $f = \lim_n f_n$ almost everwhere.
We can show that $(g+f_n)$ and $(g-f_n)$ are sequences of non-negative measurable functions.
How can we show this for $(g-f_n)$? We do get that  $f_n\le g$, but what if $f_n(x)$ is infinite for some $x$?
Then by Fatou's lemma, we have that $\int\liminf(g+f_n)dx\le\liminf\int(g+f_n)dx$. Now from here, we can obtain that
$$\int(g+f)dx \le \int gdx+\liminf \int f_ndx$$
How? How do we get the left hand side in this? I can show that $\int\liminf f_ndx = \int fdx$, but how to prove that $\liminf f_n$ is integrable to prove the former? Also I know that $\liminf(g+f_n) \geq\liminf g+\liminf(f_n)$... but how do we get the left hand side? Can I integrate it throughout but why are all these limit inferior integrable?
 A: I assume there are three major doubts. $\\$
1) What happens when $f_n(x)$ is infinite for some $x$? $\\$
2) Why $ \liminf \int (g+f_n)d\mu=\int g d\mu + \liminf f_nd\mu ? \\$ 
3) Why $ \liminf f_n $ is integrable? 
$\\$
Let's consider them one by one. $\\$
1) Note that $f_n$ is dominated by $g$ which is in $L^1$ implying that $f_n$ is in $L^1$, hence, is finite a.e. So, even if $f_n$ is infinite at some point, it doesn't matter because outside a measure zero set, $f_n$ is finite for every $n$ and measure zero sets are killed during integration. $\\$
2) Integration and limit are linear. Since $\liminf$ is actually limit in this case, so, 
$ \liminf \int (g+f_n)d\mu=\liminf \int g d\mu + \liminf f_nd\mu \\$ . Since first integral with $g$ is independent of $n$. Hence, $\liminf \int g d\mu= \int g d\mu \\$
3) First note that $\liminf f_n$ is measurable because each $f_n$ is measurable and since each $f_n$ is bounded by $g$, hence, $\liminf f_n = \sup_{n\in \mathbb{N}} \inf_{m\geq n} (f_m) \leq g $  and therefore is integratable as $g$ is integrable.
