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

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 $$f \in L^1$$?

$$f$$ 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.