2:The Riemann-Stieltjes Delta Integral
- Page ID
- 207600
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Consider a partition \(P=\{t_0, t_1, \ldots, t_n\}\in \mathcal{P}(I)\). For any \(j\in \{1, \ldots, n\}\), take \(\xi_j\in I_j\).
Let \(f\) be a function on \(I\) and \(g\) be a strictly increasing function on \(I\).
The Riemann-Stieltjes sum \(R(P, f, g)\) for the function \(f\) on the interval \(I\) with respect to the function \(g\) is defined as follows
\[
R(P, f, g)=\sum\limits_{j=1}^n f(\xi_j)\Delta g_j.
\]
Let \(g\) be a strictly increasing function on \(I\).
A function \(f\) on \(I\) is said to be Riemann-Stieltjes delta integrable on the interval \(I\) with respect to the function \(g\) if there exists a real number \(J\) such that for any \(\epsilon>0\) there exists a \(\delta=\delta(\epsilon)>0\) such that for any \(P\in \mathcal{P}_\delta (I)\) one has
\[
|R(P, f, g)-J|<\epsilon
\]
independent of the choice of \(\xi_j\in I_j\), \(j\in \{1, \ldots, n\}\).
The number \(J\) is called the Riemann-Stieltjes delta integral of the function \(f\) on the interval \(I\) with respect to the function \(g\), also denoted as
\[
\mbox{RS}\int\limits_a^b f(t)\Delta g(t).
\]
The set of all Riemann-Stieltjes delta integrable functions on the interval \(I\) with respect to the function \(g\) is denoted by \(\mathcal{R}(g, I)\).
\begin{theorem}
Let \(f\) be a function on \(I\) and \(g\) be a strictly increasing function on \(I\). If \(f\in \mathcal{R}(g, I)\), then its Riemann-Stieltjes delta integral on the interval \(I\) with respect to the function \(g\) is unique.
Proof
Suppose the contrary, i.e., assume that we have two values for \(\mbox{RS}\int\limits_a^b f(t)\Delta g(t)\), namely \(J_1\) and \(J_2\). Fix \(\epsilon>0\) arbitrarily. By the definition for the Riemann-Stieltjes delta integral, it follows that there are \(\delta_1=\delta_1(\epsilon)>0\) and \(\delta_2=\delta_2(\epsilon)>0\) such that if \(P_1\in \mathcal{P}_{\delta_1}(I)\), \(P_2\in \mathcal{P}_{\delta_2}(I)\),
then one has
\[
|R(P_1, f, g)-J_1|<\frac{\epsilon}{2}
\]
and
\[
|R(P_2, f, g)-J_2|<\frac{\epsilon}{2}.
\]
Let \(\delta=\min\{\delta_1, \delta_2\}\). Then, if \(P\in \mathcal{P}_\delta (I)\), one has
\[
|R(P, f, g)-J_1|<\frac{\epsilon}{2}\]
and
\[
|R(P, f, g)-J_2|<\frac{\epsilon}{2}.
\]
Hence,
\[\begin{eqnarray*}
|J_1-J_2|&=& |J_1-R(P, f, g)+R(P, f, g)-J_2|\\ \\
&\leq& |R(P, f, g)-J_1|+|R(P, f, g)-J_2|\\ \\
&<& \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ \\
&=& \epsilon.
\end{eqnarray*}\]
Thus,
\[
|J_1-J_2|<\epsilon \quad \mbox{for any}\quad \epsilon>0.
\]
This means
\[
|J_1-J_2|=0,
\]
which says that the two values of the Riemann-Stieltjes delta integral must be equal to each other. This completes the proof.
Let \(\mathbb{T}\), \(f\) and \(g\) be as in the third example of "The Darboux-Stieltjes Integral". Let also,
\[\begin{eqnarray*}
\xi_j&\in& I_j\\ \\
&=& [2^{j-1-n}, 2^{j-n})_{\mathbb{T}}\\ \\
&=& 2^{j-1-n}.
\end{eqnarray*}\]
Then, using the computations in the third and fourth examples of "The Darboux-Stieltjes Integral", we get
\[\begin{eqnarray*}
R(P, f, g)&=& \sum\limits_{j=1}^n f(\xi_j)\Delta g_j\\ \\
&=& \sum\limits_{j=1}^n 2^{j-1-n}\left(3\cdot 2^{2j-2n-2}\right)\\ \\
&=& \frac{3}{8}\sum\limits_{j=1}^n 2^{3j-3n}\\ \\
&=& \frac{3}{8}2^{-3n}\sum\limits_{j=1}^n 2^{3j}\\ \\
&=& \frac{3}{8}8^{-n}8\frac{8^n-1}{8-1}\\ \\
&=& \frac{3}{7}\left(1-8^{-n}\right).
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
\lim\limits_{n\to \infty}R(P, f, g)&=& \lim\limits_{n\to \infty} \frac{3}{7}\left(1-8^{-n}\right)\\ \\
&=& \frac{3}{7}.
\end{eqnarray*}\]
Thus,
\[
\mbox{RS}\int\limits_0^1 t \Delta t^2=\frac{3}{7}.
\]