Comparison Theorem Convergent Case Proof

Theorem

• Let $f(x)$, $g(x)$, $h(x)$ be cont on interval $[a,b]$
• If $\forall x\in [a,b],0\le f(x)\le g(x)$
• If ${\int }_a^{b}g(x)dx$ is cont
• Then, ${\int }_a^{b}f(x)dx$ is cont

Proof

1. Suppose $\forall x\in [a,\infty ],0\le f(x)\le g(x)$
2. Suppose ${\int }_a^{\infty }f(x)dx$ Converges
3. We want to show ${\int }_a^{\infty }f(x)dx$ Converges
4. Let $A\in [a,\infty )$ be arbitrary
5. $⟹\forall x\in [a,A]\subset [a,\infty ),0\le f(x)\le g(x)$
6. $⟹{\int }_a^{A}0dx\le {\int }_a^{A}f(x)dx\le {\int }_a^{A}g(x)dx$ as integrals preserve inequalities
7. $⟹0\le {\int }_a^{A}f(x)dx\le {\int }_a^{A}g(x)dx$
8. $⟹{\lim }_{A\to \infty }0\le {\lim }_{A\to \infty }{\int }_a^{A}f(x)dx\le {\lim }_{A\to \infty }{\int }_a^{A}g(x)dx$ as limits preserve non-strict inequalities
9. Note that ${\lim }_{A\to \infty }{\int }_a^{A}g(x)dx$ converges by (3)
10. Note that ${\lim }_{A\to \infty }{\int }_a^{A}f(x)dx$ is the area accumulation function
11. By FToC Part 2, we know that ${\lim }_{A\to \infty }{\int }_a^{A}f(x)dx$ is continuous
12. Thus, ${\int }_a^{\infty }f(x)dx$ converges