Suppose $f_n,f,g$ are regulated functions on $[0,R]$ for every $R>0$ and satisfy:
(i)$f_n \to f$ unfiformly on $[0,R]$, (ii) $|f_n(x)| \leq g(x)$ on $[0,\infty)$, and (iii) $\int^{\infty}_0g$ is convergent
Show that the improper integral $\int^{\infty}_0f$ exists and $\int^{\infty}_0f_n\to\int^{\infty}_0 $ as $n\to\infty$
A regulated function $f$ on $[a,b]$ is a function for which there exists a step function on $[a,b]$ whiih converges uniformly to $f$
So, my attempt is:
$f\in R[a,b] \Rightarrow \exists \phi_n \in S[0,R]$ such that $||f-\phi_n||_{\infty}\to 0$
So we have $f_n\to f$ uniformly $\Rightarrow ||f_n-f||_{\infty}0 \Rightarrow 0 \leq ||f_n-\phi_n||_{\infty}\leq ||f_n-f||_{\infty}+||f-\phi_n||_{\infty}\to 0 \Rightarrow f_n\to\phi_n$ uniformly.
Furthermore, $g\in R[0,R] \Rightarrow \exists \psi_n \in S[0,R] : ||g-\psi_n||_{\infty}\to 0$
$f_n\leq g \Rightarrow$ we may choose our $\phi_n$ and $\psi_n$ above such that $\phi_n \leq \psi_n$.
With such choices of $\phi_n$ and $\psi_n$, we have that
$\int^{\infty}_0g $ converges $\Rightarrow |\int^{\infty}_0|=|\lim_{n\to\infty}\int^{\infty}_0\phi_n|\leq|\lim\int^{\infty}_0\psi_n|=|\int^{\infty}_0g|<\infty$ $\Rightarrow \int^{\infty}_0f$ exists
Finally, $\lim_{n\to\infty}\int^{\infty}_0f_n=\lim_{n\to\infty}\int^{\infty}_0\phi_n=\lim_{n\to\infty}\int^{\infty}_0f$
So $\int^{\infty}_0f_n\to\int^{\infty}_0f$ as $n\to\infty$
Is this correct? Any help is appreciated!
I can see a couple of things that are not obvious to me.
Why can we choose $\phi_n$ and $\psi_n$ so that $\phi_n \leq \psi_n$ ?
You have assumed but not stated that $|\lim_{n\to\infty}\int^{\infty}_0\phi_n|=|\int^{\infty}_0\lim_{n\to\infty}\phi_n|$. Do you know that all the integrals in the sequence $|\int^{\infty}_0\phi_n|$ exist?
You may be able to answer these by choosing suitable subsequences.
I would be in favour of proving convergence by showing that $|\int^{x}_0\phi_{mn}-a|<\epsilon$ for some $a$ whenever $x$, $m$ and $n$ are sufficiently large, where $\phi_{mn}\to f_n$ as $m\to\infty$, $x$ is real and $m$, $n$ are integers.
Can $\phi_{nn}$ (my notation) be the same as $\phi_n$ (your notation)? If so the question is about changing the order of $x$, $m$, and $n$ being taken to the limit.
By the way, when I see $x_n\to y$ I assume $y$ is fixed, so $f_n\to\phi_n$ should be $f_n-\phi_n\to 0$. There are a couple of other typos too.