The Squares of R-S Integrable Functions with Increasing Integrators

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)

\begin{align} \quad \biggr \lvert \int_a^b f(x) \: d \alpha (x) \biggr \rvert \leq \int_a^b \mid f(x) \mid \: d \alpha (x) \end{align}

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:

(2)

\begin{align} \quad \sum_{k=1}^{n} [M_k(f) - m_k(f)] \Delta \alpha_k = U(P, f, \alpha) - L(P, f, \alpha) < \epsilon_1 = \frac{\epsilon}{2M} \quad (*) \end{align}

• We note that:

(3)

\begin{align} \quad M_k(f^2) = \sup \{ [f(x)]^2 : x \in [x_{k-1}, x_k] \} = \left ( \sup \{ \mid f(x) \mid : x \in [x_{k-1}, x_k] \} \right )^2 = [M_k (\mid f \mid)]^2 \end{align}

(4)

\begin{align} \quad m_k(f^2) = \inf \{ [f(x)^2] : x \in [x_{k-1}, x_k] \} = \left ( \inf \{ \mid f(x) \mid : x \in [x_{k-1}, x_k] \} \right )^2 = [m_k (\mid f \mid)]^2 \end{align}

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

(5)

\begin{align} \quad M_k(f^2) - m_k(f^2) = M_k(\mid f \mid)^2 - m_k(\mid f \mid)^2 = [M_k(\mid f \mid) + m_k(\mid f \mid)][M_k(\mid f \mid) - m_k(\mid f \mid)] < 2M[M_k(\mid f \mid) - m_k(\mid f \mid)] \end{align}

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

(6)

\begin{align} \quad U(P, f^2, \alpha) - L(P, f^2, \alpha) = \sum_{k=1}^{n} [M_k(f^2) - m_k(f^2)] \Delta \alpha_k < \sum_{k=1}^{n} 2M[M_k(\mid f \mid) - m_k (\mid f \mid)] \Delta \alpha_k = 2M \sum_{k=1}^{n} [M_k(\mid f \mid) - m_k(\mid f \mid)] \Delta \alpha_k \end{align}

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

(7)

\begin{align} \quad U(P, f^2, \alpha) - L(P, f^2, \alpha) < ... = 2M[U(P, \mid f \mid, \alpha) - L(P, \mid f \mid, \alpha)] \leq 2M[U(P, f, \alpha) - L(P, f, \alpha)] < 2M \epsilon_1 = 2M \frac{\epsilon}{2M} = \epsilon \end{align}

• 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)

\begin{align} \quad U(P, f, x) - L(P, f, x) = \sum_{k=1}^{n} [M_k(f) - m_k(f)] \Delta \alpha_k = \sum_{k=1}^{n} 2 \Delta \alpha_k = 2(1 - 0) = 2 \end{align}

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 $[0, 1]$.