# The Squares of Riemann-Stieltjes Integrable Functions with Increasing Integrators

Recall from The Absolute Value of Riemann-Stieltjes Integrals 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 $\mid f \mid$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and furthermore:

(1)Now suppose once again that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ (where $\alpha$ is an increasing function). It would be nice to know whether or not the function $f^2$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. Fortunately it is, and we can prove it by using the theorem above and Riemann's condition.

Theorem 1: Let $f$ be a function defined on $[a, b]$ and $\alpha$ be an increasing function on $[a, b]$. If $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $f^2$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. |

**Proof:**Let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ where $\alpha$ is an increasing function. Let $M > 0$ be any upper bound of $\mid f \mid$ on $[a, b]$. By Riemann's condition, for $\epsilon_1 = \frac{\epsilon}{2M} > 0$ there exists a partition $P_{\epsilon_1} \in \mathscr{P}[a, b]$ such that if $P$ is finer than $P_{\epsilon_1}$ then:

- We note that:

- Hence, if $P_{\epsilon} = P_{\epsilon_1}$, then for $P$ finer than $P_{\epsilon}$ we have that $(*)$ holds and so:

- Multiplying by $\Delta \alpha_k$ and taking the sum from $k = 1$ to $k = n$ gives us that:

- From The Absolute Value of Riemann-Stieltjes Integrals with Increasing Integrators page we see that then:

- So for all $\epsilon > 0$ there exists a partition $P_{\epsilon}$ such that if $P$ is finer than $P_{\epsilon}$ then $U(P, f^2, \alpha) - L(P, f^2, \alpha) < \epsilon$, so Riemann's condition is satisfied and $f^2$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. $\blacksquare$

It is very important to note that the converse of Theorem 1 is not true in general.

For example, consider the function $f$ defined for all $x \in [0, 1]$ by $f(x) = \left\{\begin{matrix} 1 & \mathrm{if} \: x \: \mathrm{is \: irrational}\\ -1 & \mathrm{if} \: x \: \mathrm{is \: rational} \end{matrix}\right.$ and let $\alpha (x) = x$. Then $f^2(x) = 1$ on all of $[0, 1]$ which we already know is Riemann-Stieltjes integrable from the Riemann-Stieltjes Integrals with Constant Integrands page.

However, $f$ itself is not Riemann-Stieltjes integrable. If $P = \{ 0 = x_0, x_1, ..., x_n = 1 \} \in \mathscr{P}[0, 1]$ is any partition, then for all $k \in \{ 1, 2, ..., n \}$ we have that $M_k(f) = \sup \{ f(x) : x \in [x_{k-1}, x_k] \} = 1$ and $m_k (f) = \inf \{ f(x) : x \in [x_{k-1}, x_k] \} = -1$ since every subinterval $[x_{k-1}, x_k]$ contains both rational and irrational numbers. Therefore:

(8)But $U(P, f, x) - L(P, f, x) = 2$ for all partitions $P$, so for $\epsilon_1 = 1 > 0$ we have that there exists an positive epsilon such that for all partitions $P \in \mathscr{P}[0, 1]$ we have that $U(P, f, x) - L(P, f, x) > \epsilon_1$, so $f$ does not satisfy Riemann's condition and is hence not Riemann-Stieltjes integrable with respect to $\alpha$ on $[0, 1]$.