The Product of R-S Integrable Functions with Increasing Integrators

# The Product 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 $[a, b]$.

We will now use this important theorem to show that if $f$ and $g$ are both functions defined on $[a, b]$, $\alpha$ is increasing on $[a, b]$, and $f$ 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 $[a, b]$.

 Theorem 1: Let $f$ and $g$ both be functions defined on $[a, b]$ and let $\alpha$ be an increasing function on $[a, b]$. If $f$ and $g$ are Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $fg$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.
• Proof: Let $f$ and $g$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. We note that:
(1)
\begin{align} \quad [f(x) + g(x)]^2 &= [f(x)]^2 + 2f(x)g(x) + [g(x)]^2 \\ \quad 2f(x)g(x) &= [f(x)]^2 + [g(x)]^2 - [f(x) + g(x)]^2 \\ \quad f(x)g(x) &= \frac{1}{2} [f(x)]^2 + \frac{1}{2}[g(x)]^2 - \frac{1}{2}[f(x) + g(x)]^2 \end{align}
• Therefore we have that:
(2)
\begin{align} \quad \int_a^b f(x)g(x) \: d \alpha (x) = \frac{1}{2} \int_a^b [f(x)]^2 \: d \alpha (x) + \frac{1}{2} \int_a^b [g(x)]^2 \: d \alpha (x) - \frac{1}{2} \int_a^b [f(x) + g(x)]^2 \: d \alpha x \end{align}
• Since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ we have that $\int_a^b [f(x)]^2 \: d \alpha (x)$ exists. Similarly, since $g$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ we have that $\int_a^b [g(x)]^2 \: d \alpha (x)$.
• Hence all three Riemann-Stieltjes integrals on the righthand side exist, so $\int_a^b f(x)g(x) \: d \alpha (x)$ exists, i.e., $fg$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. $\blacksquare$