3: Correspondence between the Darboux-Stieltjes Delta Integral and the Riemann-Stieltjes Delta Integral

Theorem \(\PageIndex{8}\)

Let \(f\) be a bounded function on \(I\) and \(g\) be a strictly increasing function on \(I\). Then \(f\in \mathcal{R}(g, I)\) if and only if \(f\in \mathcal{D}(g, I)\). Moreover, the value of Riemann-Stieltjes and Darboux-Stieltjes delta integrals coincide and this is denoted by
\[
\int\limits_a^b f(t)\Delta g(t).
\]

Proof

We will prove the implications separately.

1. Suppose that \(f\in \mathcal{D}(g, I)\). Fix \(\epsilon>0\) arbitrarily. By the \(\epsilon\)-criterion for Darboux-Stieltjes delta integrability, it follows that there is a \(\delta=\delta(\epsilon)>0\) such that if \(P\in \mathcal{P}_\delta(I)\), then
\[
U(P, f, g)-L(P, f, g)<\epsilon.
\]
Clearly, we have
\[\begin{equation*}
\label{17} L(P, f, g)\leq R(P, f, g)\leq U(P, f, g).
\end{equation*}\]
Then
\[\begin{eqnarray*}
R(P, f, g)&\leq& U(P, f, g)\\ \\
&<& \epsilon+ L(P, f, g)\\ \\
&\leq& \epsilon+\mbox{DS}\int\limits_a^b f(t)\Delta g(t)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
R(P, f, g)&\geq& L(P, f, g)\\ \\
&>& U(P, f, g)-\epsilon\\ \\
&\geq& \mbox{DS}\int\limits_a^b f(t)\Delta g(t)-\epsilon.
\end{eqnarray*}\]
Consequently
\[
\left|R(P, f, g)-\mbox{DS}\int\limits_a^b f(t)\Delta g(t)\right|<\epsilon
\]
for any \(\epsilon>0\). This means that
\[
\mbox{RS}\int\limits_a^b f(t)\Delta g(t)=\mbox{DS}\int\limits_a^b f(t)\Delta g(t).
\]

2. Now, suppose that \(f\in \mathcal{R}(g, I)\). Fix \(\epsilon>0\) arbitrarily. Then there is a \(\delta=\delta(\epsilon)>0\) such that
\[
\left|\mbox{RS}\int\limits_a^b f(t)\Delta g(t)-R(P, f, g)\right|<\epsilon.
\]
Let
\[
P=\{t_0, t_1, \ldots, t_n\}.
\]
Let also,
\[
m_j=\inf\limits_{t\in I_j}f(t)\quad \mbox{and}\quad M_j=\sup\limits_{t\in I_j}f(t),\quad j\in \{1, \ldots, n\}.
\]
By the definitions of \(m_j\), \(j\in \{1, \ldots, n\}\), it follows that there exists a \(\xi_j\in I_j\) so that
\[
f(\xi_j)<m_j+\epsilon,\quad j\in \{1, \ldots, n\}.
\]
Then
\[\begin{eqnarray*}
R(P, f, g)&=& \sum\limits_{j=1}^n f(\xi_j) \Delta g_j\\ \\
&<& \sum\limits_{j=1}^n (m_j+\epsilon)\Delta g_j\\ \\
&=& \sum\limits_{j=1}^n m_j\Delta g_j+\epsilon \sum\limits_{j=1}^n \Delta g_j\\ \\
&=& L(P, f, g)+\epsilon(g(b)-g(a)).
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
\underline{\int\limits_a^b} f(t)\Delta g(t)&\geq& L(P, f, g)\\ \\
&>& R(P, f, g)-\epsilon (g(b)-g(a)).
\end{eqnarray*}\]
By the definitions of \(M_j\), \(j\in \{1, \ldots, n\}\), it follows that there exists a \(\eta_j\in I_j\) so that
\[
f(\eta_j)>M_j-\epsilon,\quad j\in \{1, \ldots, n\}.
\]
Then
\[\begin{eqnarray*}
R(P, f, g)&=& \sum\limits_{j=1}^n f(\eta_j) \Delta g_j\\ \\
&>& \sum\limits_{j=1}^n (M_j-\epsilon)\Delta g_j\\ \\
&=& \sum\limits_{j=1}^n M_j\Delta g_j-\epsilon \sum\limits_{j=1}^n \Delta g_j\\ \\
&=& U(P, f, g)-\epsilon(g(b)-g(a)).
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
\overline{\int\limits_a^b} f(t)\Delta g(t)&\leq& U(P, f, g)\\ \\
&<& R(P, f, g)+\epsilon (g(b)-g(a)).
\end{eqnarray*}\]
Therefore
\[\begin{eqnarray*}
R(P, f, g)-\epsilon (g(b)-g(a))&<& \underline{\int\limits_a^b} f(t)\Delta g(t)\\ \\
&\leq& \overline{\int\limits_a^b}f(t)\Delta g(t)\\ \\
&<& R(P, f, g)+\epsilon (g(b)-g(a))
\end{eqnarray*}\]
for any \(\epsilon>0\). This means that
\[\begin{eqnarray*}
R(P, f, g)&\leq& \underline{\int\limits_a^b} f(t)\Delta g(t)\\ \\
&\leq& \overline{\int\limits_a^b}f(t)\Delta g(t)\\ \\
&\leq& R(P, f, g)
\end{eqnarray*}\]
and \(f\in \mathcal{D}(g, I)\). This completes the proof.

Theorem \(\PageIndex{9}\)

Let \(g\) be a strictly increasing function on \(I\). We have
\[\begin{equation}
\label{20}
\int\limits_a^b c \Delta g(t)=c(g(b)-g(a)),
\end{equation}\]
where \(c\) is a constant.

Proof

Let \(P=\{t_0, t_1, \ldots, t_n\}\in \mathcal{P}(I)\). Then
\[\begin{eqnarray*}
M_j&=& \sup\limits_{t\in I_j} f(t)\\ \\
&=& c,\\ \\
m_j&=& \inf\limits_{t\in I_j}f(t)\\ \\
&=& c,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
U(P, f, g)&=& \sum\limits_{j-=1}^n M_j \Delta g_j\\ \\
&=& c\sum\limits_{j=1}^n \Delta g_j\\ \\
&=& c(g(b)-g(a))
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
L(P, f, g)&=& \sum\limits_{j-=1}^n m_j \Delta g_j\\ \\
&=& c\sum\limits_{j=1}^n \Delta g_j\\ \\
&=& c(g(b)-g(a)).
\end{eqnarray*}\]
Note that there exists \(\int\limits_a^b f(t)\Delta g(t)\) and \eqref{20} holds. This completes the proof.

Theorem \(\PageIndex{10}\)

Let \(f\) be a function on \(I\) and \(g\) be a strictly increasing function on \(I\). Let also, \(t\in \mathbb{T}\) and \(f: \mathbb{T}\to \mathbb{R}\). Then
\[\begin{equation}
\label{22} \int\limits_t^{\sigma(t)} f(t)\Delta g(t)=f(t)(g(\sigma(t))-g(t)).
\end{equation}\]

Proof

If \(\sigma(t)=t\), then the assertion is evident. Let \(\sigma(t)>t\). Then
\[
\mathcal{P}([t, \sigma(t)])=\{t, \sigma(t)\}.
\]
We have
\[\begin{eqnarray*}
M&=& \sup\limits_{s\in [t, \sigma(t))}f(s)\\ \\
&=& f(t)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
m&=& \inf\limits_{s\in [t, \sigma(t))}f(s)\\ \\
&=& f(t).
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
U(P, f, g)&=& f(t)(g(\sigma(t))-g(t))\\ \\
&=& L(P, f, g).
\end{eqnarray*}\]
Note that \(\int\limits_t^{\sigma(t)}f(t)\Delta g(t)\) exists and \eqref{22} holds. This completes the proof.

Theorem \(\PageIndex{11}\)

Let \(g\) be a strictly increasing function on \(I\). If \(f: I\to \mathbb{R}\) is a Riemann-Stieltjes delta integrable function, then \(f\) is bounded.

Proof

Suppose the contrary, i.e., let \(f: I\to \mathbb{R}\) be a Riemann-Stieltjes delta integrable on \(I\) and unbounded on \(I\). Take \(\epsilon=1\). Then, using that \(f\) is Riemann-Stieltjes delta integrable on \(I\), there exist a \(\delta>0\) and a partition \(P\in \mathcal{P}_\delta(I)\) such that
\[
\left|R(P, f, g)-\int\limits_a^b f(t)\Delta g(t)\right|<1.
\]
From here, we find
\[\begin{equation}
\label{21}
\begin{array}{lll}
|R(P, f, g)|&=& \left|R(P, f, g)-\int\limits_a^b f(t)\Delta g(t)
+\int\limits_a^b f(t)\Delta g(t)\right|\\ \\
&\leq& \left|R(P, f, g)-\int\limits_a^b f(t)\Delta g(t)
\right|+\left|\int\limits_a^b f(t)\Delta g(t)
\right|\\ \\
&<& 1+\left|\int\limits_a^b f(t)\Delta g(t)
\right.
\end{array}
\end{equation}\]
Let \(P=\{t_0, t_1, \ldots, t_n\}\). Since \(f\) is unbounded on \(I\), then there exist at least one subintreval of \(P\) where the function \(f\) is unbounded. Without loss of generality, assume that it is \(I_1=[t_0, t_1]\). Take \(p_j\in I_j\), \(j\in \{2, 3, \ldots, n\}\). Since \(f\) is unbounded on \(I_1\), there is a \(p_1\in I_1\) such that
\[
|f(p_1)|>\frac{\left|\int\limits_a^b f(t)\Delta g(t)
\right|+1+\sum\limits_{j=2}^n |f(p_j)|\Delta g_j}{\Delta g_1}.
\]
Hence,
\[
R(P, f, g)=f(p_1)\Delta g_1+\sum\limits_{j=2}^n f(p_j)\Delta g_j
\]
and applying the reverse triangle inequality, we get
\[\begin{eqnarray*}
|R(P, f, g)|&=& \left|f(p_1)\Delta g_1+\sum\limits_{j=2}^n f(p_j)\Delta g_j\right|\\ \\
&\geq& |f(p_1)|\Delta g_1-\sum\limits_{j=2}^n |f(p_j)|\Delta g_j\\ \\
&>& \left|\int\limits_a^b f(t)\Delta g(t)
\right|+1+\sum\limits_{j=2}^n |f(p_j)|\Delta g_j-\sum\limits_{j=2}^n |f(p_j)|\Delta g_j\\ \\
&=& 1+\left|\int\limits_a^b f(t)\Delta g(t)
\right|.
\end{eqnarray*}\]
This contradicts with \eqref{21}. Therefore \(f\) is bounded on \(I\). This completes the proof.