# Higher Powers of Riemann-Stieltjes Integrable Functions with Increasing Integrators

Recall from The Squares of Riemann-Stieltjes Integrable Functions with Increasing Integrators page that if $f$ is a function defined on $[a, b]$ and $\alpha$ is an increasing function on $[a, b]$ then if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $f^2$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.

Furthermore, on The Product of Riemann-Stieltjes Integrable Functions with Increasing Integrators page we saw that if $f$ and $g$ are both defined on $[a, b]$ and $\alpha$ is an increasing function then if both $f$ and $g$ are Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then their product $fg$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.

In the following theorem, we will see use both results above to show that any positive power of $f$ will also be a Riemann-Stieltjes integrable function with respect to $\alpha$ on $[a, b]$.

Theorem 1: Let $f$ be a function defined on $[a, b]$ and let $\alpha$ be an increasing function on $[a, b]$. If $f$ is Riemann-Stieltjes integrable with respect to $[a, b]$, then for all $n \in \mathbb{N}$, $f^n$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. |

*We carry out the proof of Theorem 1 by induction.*

**Proof:**Let $S(k)$ for $k \geq 1$ be the statement that $f$ is Riemann-Stieltjes (which we abbreviate as "R-S" from now on) integrable with respect to $\alpha$ on $[a, b]$.

- For the base step $k = 1$ we have that $f^1 = f$ is given as R-S integrable with respect to $\alpha$ on $[a, b]$, and for $k = 2$ we have already proven (form the theorem restated above) that $f^2$ is R-S integrable with respect to $\alpha$ on $[a, b]$.

- For $k > 2$ assume that $S(k)$ is true, that is, assume that $f^k$ is R-S integrable with respect to $\alpha$ on $[a, b]$. We then want to show that $S(k+1)$ is true, i.e., show that $f^{k+1}$ is R-S integrable with respect to $\alpha$. Note that:

- We know that $f$ and $f^k$ are both R-S integrable with respect to $\alpha$ on $[a, b]$ so their product $f^{k+1}$ is also R-S integrable with respect to $\alpha$ on $[a, b]$, so $S(k+1)$ is true.

- So for all $n \in \mathbb{N}$ we have that $f^n$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. $\blacksquare$