I'll attempt to prove the "easy" part of this proof that wasn't answered here.
Any feedback on the proof itself or proof writing style would be much appreciated!
$\{f_n\}$ is a sequence of functions and $p_n$ is an increasing sequence such that $p_n \rightarrow +\infty$. We have that
- The sequence of functions $\{f_n\}$ converges uniformly to $f$ on $[a, b]$ for every $b \ge a$.
- Each $f_n$ is Riemann-integrable on $[a, b]$ for every $b \ge a$.
- $|f_n(x)| < g(x)$ almost everywhere on $[a, +\infty)$ for some nonegative g, which is improper Riemann integratable on $[a, +\infty)$.
Prove that $f$, $|f| \in \mathcal{R}([a, +\infty))$.
Please disregard (1), see correction at the bottom. We will use the Lebesgue Dominated Convergence Theorem to show that $\forall b\ge a$, $$ \lim_{n \rightarrow \infty} \int_a^b f_n = \int_a^b f \tag{1} $$
This follows because $f_n {^\rightarrow_\rightarrow} f$ and $|f_n| \le g$ a.e. on $[a,b]$, where $0 \le g \in \mathcal{R}([a,\infty))$.
Now because $f_n {^\rightarrow _\rightarrow} f$, $\forall \epsilon > 0~~\exists N \in \mathbb{N}$ such that $\forall n \ge N$ and $\forall x \in [a,b]$, $$ |f_n(x) - f(x)| < \epsilon $$
But then \begin{align} & ||f_n(x)| - |f(x)|| \le |f_n(x) - f(x)| < \epsilon \\ & \Rightarrow |f_n| {^\rightarrow_\rightarrow} |f| \tag{2} \end{align}
And given (1), (2), and 3 above, we now have $\forall b \ge a$ $$ \int_a^b |f| = \lim_{n \rightarrow \infty} \int_a^b |f_n| \le \int_a^b g \le \int_a^\infty g < \infty \tag{3} $$
The claim follows from (1) and (3). $\square$
Update: Correction for (1) -- thanks to Maksim!
We have $f \in \mathcal{R}([a,b])$ $\forall b \ge a$, because $f_n{^\rightarrow_\rightarrow} f$ and $f_n \in \mathcal{R}([a,b])$.
