Assumption that measure of set is finite necessary for showing integrable, converges in integral?

Question

Suppose $\mu(X) < \infty$, $f_n$ converges to $f$ uniformly, and each $f_n$ is integrable. I can show that (1) $f$ is integrable and (2) $\int f_n \to \int f$.

Question. Is the condition $\mu(X) < \infty$ necessary in showing (1) and (2)?

Answer 1

Yes, it's necessary:

For ($1$), let $X=[1,\infty)$ with Lebesgue measure, $f_n(x)=\frac{1}{x}1_{[1,n)}(x)$, and $f(x)=\frac{1}{x}$. Then $f_n$ is integrable and $f_n\to f$ uniformly since $0\leq f(x)-f_n(x)\leq \frac{1}{n}$, but $f$ is not integrable.

For ($2$), if $X=\mathbb{R}$ with Lebesgue measure, $f_n=\frac{1}{n}1_{[0,n]}$, and $f=0$, then $f_n\to f$ uniformly, but $\int f_n=1$ for all $n$ while $\int f=0$.