I am a bit rusty on formally proving the existence of improper integrals. I did a somewhat simple proof, but it feels sketchy and incomplete. I was hoping someone could verify it.
Problem: Suppose $f,g:(0,\infty) \rightarrow \mathbb{R}$ satisfy the following:
- $g,f \in \mathcal{R}$ (Riemann-integrable) on $[t,T]$ for any $0<t<T<\infty$,
- $|f| \leq g$,
- $\int_0^\infty g(x)dx$ converges.
Show $\int_a^\infty f(x)dx$ converges for any $a>0$.
Proof: Let $a>0$. Choose $N \in \mathbb{Z}_+$ with $a \leq N$. Remark $0 \leq |f| \leq g$ implies $\int_t^T g(x)dx \geq 0$ for any $0<t<T<\infty$. Also note $\int_0^\infty g(x)dx = \int_0^N g(x)dx + \int_N^\infty g(x)dx$. Since $\int_0^\infty g(x)dx$ converges and $0 \leq |f| \leq g$, then $\int_N^\infty g(x)dx$ must converge. By the integral test, it follows $\sum_{n=N}^\infty g(n)$ converges. Since $|f| \leq g$, it follows $\sum_{n=N}^\infty |f(n)|$ converges. Therefore $\sum_{n=N}^\infty f(n)$ converges, and so $\int_N^\infty f(x)dx$ exists. Note $\int_a^\infty f(x)dx = \int_a^N f(x)dx + \int_N^\infty f(x)dx$. Recall $f \in \mathcal{R}$ on [t,T] for any $0<t<T<\infty$. Then $\int_a^N f(x)dx$ exists. Therefore $\int_a^\infty f(x)dx = \int_a^N f(x)dx + \int_N^\infty f(x)dx$ converges. $\square$
Is this proof good? Are there any incorrect statements or jumps of logic? Thank you.