Let $f(x, n): \ [0, \infty) \rightarrow \mathbb{R}$ and $f(x): \ [0, \infty) \rightarrow \mathbb{R}$ be functions such that $\lim_{n \to \infty} f(x,n)=f(x)$ somehow. Let's further assume that the improper Riemann integrals $\int_0^{\infty}f(x, n)$ and $\int_0^{\infty} f(x)$ all converge. I want to show that: $$ \lim_{n \to \infty} \int_{0}^{\infty} f(x,n)\mathrm{d}x = \int_0^{\infty} f(x) \mathrm{d}x $$ Under some conditions. If it was a proper integral over a compact set, ($ \int_a^b f(x,n)$), it would be easy to show that the condition: "$f(x, n)$ converges to $f(x)$ uniformly in $[a,b]$" suffices. However, I haven't succeed to replicate the case for the improper case.
So now I wonder what condition suffices. Does "uniform convergence in $[0, \infty)$" suffice? What about "uniform convergence on any compact subset"? Maybe other kinds of conditions?
Thank you.
This is formally encapsulated by the Dominated Convergence theorem, which says that if there is a $g(x)$ such that $|f(x,n)|\leq g(x)$ and $g(x)$ is integrable then the limit can pass through the integral. Otherwise, take $f(x,n)$ to be a bump with fixed area 1 (and zero elsewhere), that moves off to infinity as $n$ gets larger. Then $f(x)=0$, but the limit of the integral does not equal the integral of the limit.