6: Mean Value Theorems for the Riemann-Strieltjes Delta Integral
- Page ID
- 207604
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Let \(g\) be a strictly increasing function on \(I\), \(f\in \mathcal{R}(g, I)\) be a bounded function on \(I\) and
\[
m=\inf\limits_{t\in [a, b)} f(t)\quad \mbox{and}\quad M=\sup\limits_{t\in [a, b)} f(t).
\]
Then there is a real number \(\Lambda\in [m, M]\) such that
\[\begin{equation}
\label{66} \int\limits_a^b f(t)\Delta g(t)=\Lambda (g(b)-g(a)).
\end{equation}\]
Proof
We have
\[
m\leq f(t)\leq M,\quad t\in I.
\]
Then, we find
\[\begin{eqnarray*}
m(g(b)-g(a))&=& \int\limits_a^bm \Delta g(t)\\ \\
&\leq& \int\limits_a^b f(t)\Delta g(t)\\ \\
&\leq& M\int\limits_a^b 1\Delta g(t)\\ \\
&=& M(g(b)-g(a)).
\end{eqnarray*}\]
Hence, we get
\[
m\leq \frac{\int\limits_a^b f(t)\Delta g(t)}{g(b)-g(a)}\leq M.
\]
Set
\[
\Lambda=\frac{\int\limits_a^b f(t)\Delta g(t)}{g(b)-g(a)}.
\]
Then \(\Lambda\in [m, M]\) and \eqref{66} holds. This completes the proof.
For the next mean value theorem we have a need of the following auxiliary result.
Let \(p_j\), \(j\in \{1, \ldots, n\}\), be nonnegative real numbers such that
\[
p_1\geq p_2\geq \ldots \geq p_n\geq 0.
\]
Let also, \(q_j\), \(j\in \{1, \ldots, n\}\), be nonnegative real numbers and the real numbers
\[
S_k=\sum\limits_{j=1}^k q_j,\quad k\in \{1, \ldots, n\},
\]
satisfy the inequalities
\[
m\leq S_k\leq M,\quad k\in \{1, \ldots, n\},
\]
for some nonnegative real numbers \(m\) and \(M\). Then one has
\[
mp_1\leq \sum\limits_{j=1}^n p_j q_j\leq Mp_1.
\]
Proof
We have
\[\begin{eqnarray*}
\sum\limits_{j=1}^n p_j q_j&\geq& p_1q_1\\ \\
&=& p_1S_1\\ \\
&\geq& mp_1
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\sum\limits_{j=1}^n p_j q_j&\leq& p_1\sum\limits_{j=1}^nq_j\\ \\
&=& p_1S_n\\ \\
&\leq& Mp_1.
\end{eqnarray*}\]
This completes the proof.
Let \(g\) be a strictly increasing function on \(I\), \(f\in \mathcal{R}(g, I)\) be a nonnegative bounded function and the function
\[
F(t)=\int\limits_a^t f(s)\Delta s,\quad t\in I,
\]
satisfies the inequalities
\[
m(g(b)-g(a))\leq F(t)\leq M(g(b)-g(a)),\quad t\in I,
\]
for some nonnegative constants \(m\) and \(M\). If \(h\in \mathcal{R}(g, I)\) is a nonnegative nonincreasing function on \(I\), then there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda h(a)(g(b)-g(a)).
\]
Proof
Fix \(\epsilon>0\) arbitrarily. Since \(f, h\in \mathcal{R}(g, I)\), we get \(fh\in \mathcal{R}(g, I)\). By the \(\epsilon\)-criterion and the definition for Riemann-Stieltjes delta integral, it follows that there is a partition \(P=\{t_0, t_1, \ldots, t_n\}\in \mathcal{P}(I)\) such that
\[
U(P, f, g)-L(P, f, g)<\epsilon
\]
and
\[\begin{equation}
\label{72} \left|\sum\limits_{j=1}^nf(t_{j-1}) h(t_{j-1})\Delta g_j-\int\limits_a^b f(t)h(t)\Delta g(t)\right|<\epsilon.
\end{equation}\]
By the first inequality, we find
\[\begin{equation}
\label{71} \sum\limits_{j=1}^n (M_j-m_j)\Delta g_j<\epsilon,
\end{equation}\]
where
\[\begin{eqnarray*}
M_j&=& \sup\limits_{t\in I_j} f(t),\\ \\
m_j&=& \inf\limits_{t\in I_j}f(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Now, using that
\[
m_j\leq f(t)\leq M_j,\quad t\in I_j,\quad j\in \{1, \ldots, n\},
\]
and \(h\geq 0\) on \(I\), we find
\[\begin{equation}
\label{73}
\begin{array}{lll}
\sum\limits_{j=1}^n m_j h(t_{j-1})\Delta g_j&\leq& \sum\limits_{j=1}^n f(t_{j-1})h(t_{j-1}) \Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n M_j h(t_{j-1}) \Delta g_j.
\end{array}
\end{equation}\]
By Theorem \(\PageIndex{26}\), it follows that for any \(j\in \{1, \ldots, n\}\) there is \(\Lambda_j\in [m_j, M_j]\) such that
\[
\int\limits_{t_{j-1}}^{t_j}f(t)\Delta g(t)=\Lambda_j \Delta g_j.\]
Consider the numbers
\[S_k=\sum\limits_{j=1}^k \Lambda_j \Delta g_j,\quad k\in \{1, \ldots, n\}.
\]
We have
\[\begin{eqnarray*}
S_k&=& \sum\limits_{j=1}^k \left(\int\limits_{t_{j-1}}^{t_j}f(t)\Delta g(t)\right)\\ \\
&=& \int\limits_a^{t_k}f(t)\Delta g(t),\quad k\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
m(g(b)-g(a))&\leq& S_k\\ \\
&\leq& M(g(b)-g(a)),\quad k\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Set
\[\begin{eqnarray*}
p_j&=& h(t_{j-1}),\\ \\
q_j&=& \Lambda_j \Delta g_j,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Since \(h\) is nonincreasing on \(I\), we have
\[
p_1\geq p_2\geq \ldots \geq p_n.
\]
Thus, the numbers \(p_j, S_j, q_j\), \(j\in \{1, \ldots, n\}\), satisfy all conditions of the lemma of Abel. Therefore
\[\begin{equation}
\label{74}
\begin{array}{lll}
m(g(b)-g(a))h(a)&\leq& \sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\\ \\
&\leq& M(g(b)-g(a))h(a).
\end{array}
\end{equation}\]
On the other hand, we have the inequalities
\[\begin{eqnarray*}
\sum\limits_{j=1}^n m_j h(t_{j-1})\Delta g_j&\leq& \sum\limits_{j=1}^n \Lambda_j h(t_{j-1})\Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n M_j h(t_{j-1}) \Delta g_j.
\end{eqnarray*}\]
From the last inequalities and \eqref{73}, we find
\[\begin{eqnarray*}
\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j&\leq& \sum\limits_{j=1}^n h(t_{j-1})(M_j-\Lambda_j)\Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n h(t_{j-1}) (M_j-m_j)\Delta g_j\\ \\
&\leq& h(a)\sum\limits_{j=1}^n (M_j-m_j)\Delta g_j
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j&\geq& \sum\limits_{j=1}^n h(t_{j-1})(m_j-\Lambda_j)\Delta g_j\\ \\
&\geq& \sum\limits_{j=1}^n h(t_{j-1}) (m_j-M_j)\Delta g_j\\ \\
&\geq& h(a)\sum\limits_{j=1}^n (m_j-M_j)\Delta g_j.
\end{eqnarray*}\]
Therefore
\[
\left|\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j\right|\leq h(a)\sum\limits_{j=1}^n (M_j-m_j)\Delta g_j.
\]
Now, applying the inequality \eqref{71}, we arrive at the inequality
\[
\left|\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j\right|<\epsilon h(a).
\]
From here and \eqref{72}, we get
\[
\left|\int\limits_a^b f(t)h(t)\Delta g(t)-\sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\right|
\]
\[
\begin{eqnarray*}
&=&\bigg|\int\limits_a^b f(t)h(t)\Delta g(t)-\sum\limits_{j=1}^n f(t_{j-1})h(t_{j-1})\Delta g_j\\ \\
&&+\sum\limits_{j=1}^n f(t_{j-1}) h(t_{j-1}) \Delta g_j-\sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\bigg|\\ \\
&\leq& \left|\int\limits_a^b f(t)h(t)\Delta g(t)-\sum\limits_{j=1}^n f(t_{j-1})h(t_{j-1})\Delta g_j\right|\\ \\
&&+\left|\sum\limits_{j=1}^n f(t_{j-1}) h(t_{j-1}) \Delta g_j-\sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\right|\\ \\
&<& \epsilon+\epsilon h(a).
\end{eqnarray*}\]
Now, applying \eqref{74}, we obtain
\[\begin{eqnarray*}
\int\limits_a^b f(t)h(t)\Delta g(t)&\leq& \sum\limits_{j=1}^n h(t_{j-1})\Lambda_j \Delta g_j+\epsilon+\epsilon h(a)\\ \\
&<& M(g(b)-g(a))h(a)+\epsilon+\epsilon h(a)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\int\limits_a^b f(t)h(t)\Delta g(t)&\geq& \sum\limits_{j=1}^n h(t_{j-1})\Lambda_j \Delta g_j-\epsilon-\epsilon h(a)\\ \\
&>& m(g(b)-g(a))h(a)-\epsilon-\epsilon h(a),
\end{eqnarray*}\]
i.e.,
\[
m h(a)(g(b)-g(a))-\epsilon -\epsilon h(a)< \int\limits_a^b f(t)h(t)\Delta g(t)< Mh(a)(g(b)-g(a))+\epsilon +\epsilon h(a).
\]
Taking the limit as \(\epsilon\to 0\) in the last inequalities, we find
\[
mh(a)(g(b)-g(a))\leq \int\limits_a^b f(t)h(t)\Delta g(t)\leq Mh(a)(g(b)-g(a)).
\]
If \(h(a)=0\), then the assertion is evident. Let \(h(a)\ne 0\). Then
\[
m\leq \frac{\int\limits_a^b f(t)h(t)\Delta g(t)}{h(a)(g(b)-g(a))}\leq M.
\]
Let
\[
\Lambda=\frac{\int\limits_a^b f(t)h(t)\Delta g(t)}{h(a)(g(b)-g(a))}.
\]
Then \(\Lambda\in [m, M]\) and
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda h(a)(g(b)-g(a)).
\]
This completes the proof.
Let \(g\) be a strictly increasing function on \(I\), \(f\in \mathcal{R}(g, I)\) be a bounded nonnegative function and the function
\[
F(t)=\int\limits_a^t f(s)\Delta s,\quad t\in I,
\]
satisfies the inequalities
\[
m(g(b)-g(a))\leq F(t)\leq M(g(b)-g(a)),\quad t\in I,
\]
for some nonnegative constants \(m\) and \(M\). If \(h\in \mathcal{R}(g, I)\) is any monotone function on \(I\), then there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda (h(a)-h(b))(g(b)-g(a))+h(b)\int\limits_a^b f(t)\Delta g(t).
\]
Proof
Suppose that \(h\in \mathcal{R}(g, I)\) is any nonincreasing nonnegative function on \(I\). Define the function
\[
l(t)=h(t)-h(b),\quad t\in I.
\]
Then \(l\) is a nonincreasing function on \(I\) and
\[\begin{eqnarray*}
l(t)&\geq& l(b)\\ \\
&=& h(b)-h(b)\\ \\
&=& 0,\quad t\in I.
\end{eqnarray*}\]
Then, we can apply Theorem \(\PageIndex{27}\). Hence, there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)l(t)\Delta g(t)=\Lambda l(a)(g(b)-g(a)),
\]
or
\[
\int\limits_a^b f(t)(h(t)-h(b))\Delta g(t)= \Lambda (h(a)-h(b))(g(b)-g(a)),
\]
or
\[
\int\limits_a^b f(t)h(t)\Delta g(t)-h(b)\int\limits_a^b f(t)\Delta g(t)= \Lambda (h(a)-h(b))(g(b)-g(a)),
\]
whereupon
\[
\int\limits_a^b f(t)h(t)\Delta g(t)= \Lambda (h(a)-h(b))(g(b)-g(a))+h(b)\int\limits_a^b f(t)\Delta g(t).
\]
Now, assume that \(h\in \mathcal{R}(g, I)\) is any nondecreasing function. Define the function
\[
h_1(t)=-h(t),\quad t\in I.
\]
Then \(h_1\) is nonincreasing on \(I\). Set
\[
m(t)=h_1(t)-h_1(b),\quad t\in I.
\]
We have that \(m\) is nondecreasing on \(I\) and
\[\begin{eqnarray*}
m(t)&\geq& m(b)\\ \\
&=& h_1(b)-h_1(b)\\ \\
&=& 0,\quad t\in I.
\end{eqnarray*}\]
Thus, we can apply Theorem \(\PageIndex{27}\). Then there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)m(t)\Delta g(t)=\Lambda m(a)(g(b)-g(a)),
\]
or
\[\begin{eqnarray*}
\int\limits_a^b f(t)(h_1(t)-h_1(b))\Delta g(t)&=& \Lambda (h_1(a)-h_1(b))(g(b)-g(a))\\ \\
&=& \Lambda (-h(a)+h(b))(g(b)-g(a),
\end{eqnarray*}\]
or
\[
\int\limits_a^b f(t)(h(t)-h(b))\Delta g(t)=\Lambda (h(a)-h(b))(g(b)-g(a)),
\]
or
\[
\int\limits_a^b f(t)h(t)\Delta g(t)-h(b)\int\limits_a^b f(t)\Delta g(t)=\Lambda (h(a)-h(b))(g(b)-g(a)),
\]
whereupon
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda (h(a)-h(b))(g(b)-g(a))+h(b)\int\limits_a^b f(t)\Delta g(t).
\]
This completes the proof.
Let \(g\) be a strictly increasing function on \(I\), \(f\in \mathcal{R}(g, I)\) be a nonnegative bounded function and the function
\[
\Phi(t)=\int\limits_t^b f(s)\Delta s,\quad t\in I,
\]
satisfies the inequalities
\[
m(g(b)-g(a))\leq \Phi(t)\leq M(g(b)-g(a)),\quad t\in I,
\]
for some nonnegative constants \(m\) and \(M\). If \(h\in \mathcal{R}(g, I)\) is a nonnegative nondecreasing function on \(I\), then there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda h(b)(g(b)-g(a)).
\]
Proof
Fix \(\epsilon>0\) arbitrarily. Because \(f, h\in \mathcal{R}(g, I)\), applying we get \(fh\in \mathcal{R}(g, I)\). From the \(\epsilon\)-criterion and the definition for Riemann-Stieltjes delta integral, it follows that there is a partition \(P=\{t_0, t_1, \ldots, t_n\}\in \mathcal{P}(I)\) such that
\[
U(P, f, g)-L(P, f, g)<\epsilon\]
and
\[\begin{equation}
\label{81} \left|\sum\limits_{j=1}^n h(t_{j-1})\Delta g_j-\int\limits_a^b f(t)h(t)\Delta g(t)\right|<\epsilon.
\end{equation}\]
From the first inequality, we find
\[\begin{equation}
\label{82} \sum\limits_{j=1}^n (M_j-m_j)\Delta g_j<\epsilon,
\end{equation}\]
where
\[\begin{eqnarray*}
M_j&=& \sup\limits_{t\in I_j} f(t),\\ \\
m_j&=& \inf\limits_{t\in I_j}f(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Hence, using that
\[
m_j\leq f(t)\leq M_j,\quad t\in I_j,\quad j\in \{1, \ldots, n\},
\]
and \(h\geq 0\) on \(I\), we find
\[\begin{equation}
\label{83}
\begin{array}{lll}
\sum\limits_{j=1}^n m_j h(t_{j-1})\Delta g_j&\leq& \sum\limits_{j=1}^n f(t_{j-1})h(t_{j-1}) \Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n M_j h(t_{j-1}) \Delta g_j.
\end{array}
\end{equation}\]
By Theorem \(\PageIndex{26}\), it follows that for any \(j\in \{1, \ldots, n\}\) there is \(\Lambda_j\in [m_j, M_j]\) such that
\[
\int\limits_{t_{j-1}}^{t_j}f(t)\Delta g(t)=\Lambda_j \Delta g_j.
\]
Consider the numbers
\[
S_k=\sum\limits_{j=k+1}^n \Lambda_j \Delta g_j,\quad k\in \{1, \ldots, n\}.
\]
We have
\[\begin{eqnarray*}
S_k&=& \sum\limits_{j=k+1}^n \left(\int\limits_{t_{j-1}}^{t_j}f(t)\Delta g(t)\right)\\ \\
&=& \int\limits_{t_k}^bf(t)\Delta g(t),\quad k\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
m(g(b)-g(a))&\leq& S_k\\ \\
&\leq& M(g(b)-g(a)),\quad k\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Set
\[\begin{eqnarray*}
p_j&=& h(t_{n-j+1}),\\ \\
q_j&=& \Lambda_j \Delta g_j,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Since \(h\) is nondecreasing on \(I\), we have
\[
p_1\geq p_2\geq \ldots \geq p_n.
\]
Thus, the numbers \(p_j, S_j, q_j\), \(j\in \{1, \ldots, n\}\), satisfy all conditions of the lemma of Abel. Therefore
\[\begin{equation}
\label{84}
\begin{array}{lll}
m(g(b)-g(a))h(b)&\leq& \sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\\ \\
&\leq& M(g(b)-g(a))h(b).
\end{array}
\end{equation}\]
On the other hand, we have the inequalities
\[\begin{eqnarray*}
\sum\limits_{j=1}^n m_j h(t_{j-1})\Delta g_j&\leq& \sum\limits_{j=1}^n \Lambda_j h(t_{j-1})\Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n M_j h(t_{j-1}) \Delta g_j.
\end{eqnarray*}\]
From the last inequalities and \eqref{83}, we find
\[\begin{eqnarray*}
\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j&\leq& \sum\limits_{j=1}^n h(t_{j-1})(M_j-\Lambda_j)\Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n h(t_{j-1}) (M_j-m_j)\Delta g_j\\ \\
&\leq& h(b)\sum\limits_{j=1}^n (M_j-m_j)\Delta g_j
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j&\geq& \sum\limits_{j=1}^n h(t_{j-1})(m_j-\Lambda_j)\Delta g_j\\ \\
&\geq& \sum\limits_{j=1}^n h(t_{j-1}) (m_j-M_j)\Delta g_j\\ \\
&\geq& h(b)\sum\limits_{j=1}^n (m_j-M_j)\Delta g_j.
\end{eqnarray*}\]
Therefore
\[
\left|\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j\right|\leq h(b)\sum\limits_{j=1}^n (M_j-m_j)\Delta g_j.
\]
Now, applying the inequality \eqref{82}, we arrive at the inequality
\[
\left|\sum\limits_{j=1}^n h(t_{j-1}) (f(t_{j-1})-\Lambda_j)\Delta g_j\right|<\epsilon h(b).
\]
From here and \eqref{81}, we get
\[
\left|\int\limits_a^b f(t)h(t)\Delta g(t)-\sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\right|
\]
\[\begin{eqnarray*}
&=&\bigg|\int\limits_a^b f(t)h(t)\Delta g(t)-\sum\limits_{j=1}^n f(t_{j-1})h(t_{j-1})\Delta g_j\\ \\
&&+\sum\limits_{j=1}^n f(t_{j-1}) h(t_{j-1}) \Delta g_j-\sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\bigg|\\ \\
&\leq& \left|\int\limits_a^b f(t)h(t)\Delta g(t)-\sum\limits_{j=1}^n f(t_{j-1})h(t_{j-1})\Delta g_j\right|\\ \\
&&+\left|\sum\limits_{j=1}^n f(t_{j-1}) h(t_{j-1}) \Delta g_j-\sum\limits_{j=1}^n h(t_{j-1}) \Lambda_j \Delta g_j\right|\\ \\
&<& \epsilon+\epsilon h(b).
\end{eqnarray*}\]
Now, applying \eqref{84}, we obtain
\[\begin{eqnarray*}
\int\limits_a^b f(t)h(t)\Delta g(t)&\leq& \sum\limits_{j=1}^n h(t_{j-1})\Lambda_j \Delta g_j+\epsilon+\epsilon h(b)\\ \\
&<& M(g(b)-g(a))h(b)+\epsilon+\epsilon h(b)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\int\limits_a^b f(t)h(t)\Delta g(t)&\geq& \sum\limits_{j=1}^n h(t_{j-1})\Lambda_j \Delta g_j-\epsilon-\epsilon h(b)\\ \\
&>& m(g(b)-g(a))h(b)-\epsilon-\epsilon h(b),
\end{eqnarray*}\]
i.e.,
\[
m h(b)(g(b)-g(a))-\epsilon -\epsilon h(b)< \int\limits_a^b f(t)h(t)\Delta g(t)< Mh(b)(g(b)-g(a))+\epsilon +\epsilon h(b).
\]
Taking the limit as \(\epsilon\to 0\) in the last inequalities, we find
\[
mh(b)(g(b)-g(a))\leq \int\limits_a^b f(t)h(t)\Delta g(t)\leq Mh(b)(g(b)-g(a)).
\]
If \(h(b)=0\), then the assertion is evident. Let \(h(b)\ne 0\). Then
\[
m\leq \frac{\int\limits_a^b f(t)h(t)\Delta g(t)}{h(b)(g(b)-g(a))}\leq M.
\]
Let
\[
\Lambda=\frac{\int\limits_a^b f(t)h(t)\Delta g(t)}{h(b)(g(b)-g(a))}.
\]
Then \(\Lambda\in [m, M]\) and
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda h(b)(g(b)-g(a)).
\]
This completes the proof.
Let \(g\) be a strictly increasing function on \(I\), \(f\in \mathcal{R}(g, I)\) be a nonnegative bounded function and the function
\[
\Phi(t)=\int\limits_t^b f(s)\Delta s,\quad t\in I,\]
satisfies the inequalities
\[
m(g(b)-g(a))\leq \Phi(t)\leq M(g(b)-g(a)),\quad t\in I,
\]
for some nonnegative constants \(m\) and \(M\). If \(h\in \mathcal{R}(g, I)\) is any monotone function on \(I\), then there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda (h(b)-h(a))(g(b)-g(a))+h(a)\int\limits_a^b f(t)\Delta g(t).
\]
Proof
Suppose that \(h\in \mathcal{R}(g, I)\) is any nondecreasing function on \(I\). Define the function
\[
l(t)=h(t)-h(a),\quad t\in I.
\]
Then \(l\) is a nondecreasing function on \(I\) and
\[\begin{eqnarray*}
l(t)&\geq& l(a)\\ \\
&=& h(a)-h(a)\\ \\
&=& 0,\quad t\in I.
\end{eqnarray*}\]
Then, we can apply Theorem \(\PageIndex{29}\). Hence, there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)l(t)\Delta g(t)=\Lambda l(b)(g(b)-g(a)),
\]
or
\[
\int\limits_a^b f(t)(h(t)-h(a))\Delta g(t)= \Lambda (h(b)-h(a))(g(b)-g(a)),
\]
or
\[
\int\limits_a^b f(t)h(t)\Delta g(t)-h(a)\int\limits_a^b f(t)\Delta g(t)= \Lambda (h(b)-h(a))(g(b)-g(a)),
\]
whereupon
\[
\int\limits_a^b f(t)h(t)\Delta g(t)= \Lambda (h(b)-h(a))(g(b)-g(a))+h(a)\int\limits_a^b f(t)\Delta g(t).
\]
Now, assume that \(h\in \mathcal{R}(g, I)\) is any nonincreasing function. Define the function
\[
h_1(t)=-h(t),\quad t\in I.
\]
Then \(h_1\) is nondecreasing on \(I\). Set
\[
m(t)=h_1(t)-h_1(a),\quad t\in I.
\]
We have that $m$ is nondecreasing on \(I\) and
\[\begin{eqnarray*}
m(t)&\geq& m(a)\\ \\
&=& h_1(a)-h_1(a)\\ \\
&=& 0,\quad t\in I.
\end{eqnarray*}\]
Thus, we can apply Theorem \(\PageIndex{29}\). Then there is a number \(\Lambda\in [m, M]\) such that
\[
\int\limits_a^b f(t)m(t)\Delta g(t)=\Lambda m(b)(g(b)-g(a)),
\]
or
\[\begin{eqnarray*}
\int\limits_a^b f(t)(h_1(t)-h_1(a))\Delta g(t)&=& \Lambda (h_1(b)-h_1(a))(g(b)-g(a))\\ \\
&=& \Lambda (-h(b)+h(a))(g(b)-g(a),
\end{eqnarray*}\]
or
\[
\int\limits_a^b f(t)(h(t)-h(a))\Delta g(t)=\Lambda (h(b)-h(a))(g(b)-g(a)),
\]
or
\[
\int\limits_a^b f(t)h(t)\Delta g(t)-h(a)\int\limits_a^b f(t)\Delta g(t)=\Lambda (h(b)-h(a))(g(b)-g(a)),
\]
whereupon
\[
\int\limits_a^b f(t)h(t)\Delta g(t)=\Lambda (h(b)-h(a))(g(b)-g(a))+h(a)\int\limits_a^b f(t)\Delta g(t).
\]
This completes the proof.