Uniform convergence does not guarantee convergence of integrals when the domain has infinite measure

Question

Let $(X,\Sigma,\mu)$ be a measure space, such that $\mu(X)=\infty$.

Let $f_n:X \to \mathbb{R}$ be measurable real-valued functions, which converge uniformly to a function $f$. Suppose that $f_n \in L^1$. Is it true that $f \in L^1$? Does $\int_X f_n \to \int_X f$ hold?

Does the convergence $\int_X f_n \to \int_X f$ hold if we assume in advance that $f \in L^1$?

Answer 1

$f$ many not be in $L^1$: consider $f_n(x)=\frac1x\chi_{[1,n]}(x)$, which converges uniformly to $\frac1x\chi_{[1,\infty)}(x)$.

For the last question, consider $f_n(x)=\frac1n\sin^2(x)\chi_{[0,2n\pi]}(x)$ which converges uniformly to $0$.

Added: However, the integrals of a sequence of non-negative functions that converges to a function not in $L^1$ diverge by Fatou's lemma.