Real Analysis, Folland Problem 2.3.19 Integration of Complex Functions 
Problem 2.3.19 - Suppose $\{f_n\}\subset L^1(\mu)$ 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).

Attempted proof a.) Let $\{f_n\}\subset L^1(\mu)$, $f_n\rightarrow f$ uniformly, and $\mu(X) < \infty$. There exists an $N\in\mathbb{N}$ such that $|f_n(x) - f(x)| \leq 1$ for all $n\geq N$ and for all $x\in X$. Therefore, $$\int |f|\leq \int |f-f_N| + \int |f_N|\leq \int 1 + \int |f_N| < \infty$$
so $f\in L^1(\mu)$. Set $g = 1 + |f|$. Then $\int g = \mu(X) + \int |f|$, so $g\in L^1(\mu)$ and $|f_n|\leq g$ for all $n\geq N$. So by the Dominated Convergence Theorem $$\lim_{n\rightarrow \infty}\int f_n = \int f$$
I do not understand in this proof why $$\int 1 = \mu(X)$$ where does this come from? I also haven't figure out anything clever for part (b) yet. Any suggestions is greatly appreciated.
 A: 
Problem 2.3.19 - Suppose $\{f_n\}\subset L^1(\mu)$ 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).

Proof
Initial remark: We know from the definition of integral for simple functions that, if $E\in \mathcal{M}$ , $$\int \chi_E = \mu(E)$$
In particular, the function constant equal to $1$ coincides to $\chi_X$. So
$$\int 1 = \int \chi_X = \mu(X)$$
a.) Let $\{f_n\}\subset L^1(\mu)$, $f_n\rightarrow f$ uniformly, and $\mu(X) < \infty$. There exists an $N\in\mathbb{N}$ such that $|f_n(x) - f(x)| \leq 1$ for all $n\geq N$ and for all $x\in X$. Therefore, $$\int |f|\leq \int |f-f_N| + \int |f_N|\leq \int 1 + \int |f_N| = \mu(X) + \int |f_N|< \infty$$
so $f\in L^1(\mu)$.
Now, let $g = 1 + |f|$. Then $g \geq 0$ and $\int g = \mu(X) + \int |f|<\infty$, so $g\in L^1(\mu)$ and $|f_n|\leq g$ for all $n\geq N$. So by the Dominated Convergence Theorem $$\lim_{n\rightarrow \infty}\int f_n = \int f$$
b.) The condition $\mu(X)<+\infty$ is essential in part a.).
Consider $X=\mathbb{R}$ and $\mu$ the Lebesgue measure (defined on the Borel or Lebesgue $\sigma$-algebra).
For all $n\in \mathbb{N}$, $ n\geq 1$, let $f_n=\frac{1}{n} \chi_{[0,n]}$. It is easy to prove that
$\{f_n\}\subset L^1(\mu)$ and $f_n\rightarrow 0$ uniformly, but, for all $n$,
$\int f_n =1$. So
$$\int f_n =1 \nrightarrow 0= \int 0$$
