Let’s start shall we.
$\textbf{Problem 19.}$ Suppose $\left\{f_n\right\} \subset L^1(\mu)$ and $f_n \rightarrow f$ uniformly. a. If $\mu(X)<\infty$, then $f \in L^1(\mu)$ and $\int f_n \rightarrow \int f$.
b. If $\mu(X)=\infty$, the conclusions of (a) can fail. (Find examples on $\mathbb{R}$ with Lebesgue measure.)
$\textbf{Solution.}$ Since $f_n \rightarrow f$ uniformly, we have that there is an $N$ such that $n \ge N$ and all $x \in X$
$$|f|<1+|f_n|$$.
This implies that
$$ \int |f| < \int 1 + \int |f_n| = \mu(X) + c < \infty$$ Therefore, $f \in L^1(\mu)$. It is now easy to do the same manipulation so that $$|f_n| < 1+|f|$$ and finish off with the DCT.
$\textbf{b}$. Consider $f_n = 1_{(0,n)} n^{-1}$. $f_n$ converges uniformly to $0$. However $\mu(X)=\infty$ and $\int f_n = 1$.
$\textbf{Problem 20.}$ (A generalized Dominated Convergence Theorem) If $f_n, g_n, f, g \in L^1, f_n \rightarrow f$ and $g_n \rightarrow g$ a.e., $\left|f_n\right| \leq g_n$, and $\int g_n \rightarrow \int g$, then $\int f_n \rightarrow \int f$. (Rework the proof of the dominated convergence theorem.)
$\textbf{Solution.}$. We notice that $g_n-f_n \ge 0$ and that $f_n+g_n \ge 0$ so that both series are in $L^+(\mu)$. Therefore, by Fatou’s Lemma:
$$ \int f+g \le \liminf \int f_n + \int g$$ so that
$$ \int f \le \liminf \int f_n$$.
The same logic applies to the second series:
$$ \int g-f \le \int g -\liminf f_m = \int g+\limsup \int f_n $$
so that $$ \int f \ge \limsup \int f_n$$. The result follows.
$\textbf{21.}$ Suppose $f_n, f \in L^1$ and $f_n \rightarrow f$ a.e. Then $\int\left|f_n-f\right| \rightarrow 0$ iff $\int\left|f_n\right| \rightarrow \int|f|$. (Use Exercise 20.)
$\textbf{Solution.}$ Let $g_n=|f_n|-|f|$. Also, $g_n \in L^1(\mu)$. We then have $|g_n| \le |f_n-f|$. Also note that $\lim |f_n-f|=0$ a.e. Also by our assumption we have that $\lim \int |f_n-f| = 0$. By the generalized DCT, we have that $\lim \int g_n = \int g$ or that $\lim \int |f_n| = \int |f|$. It is easy to prove the other way in the same manner.