5: Some Inequalities for the Riemann-Stieltjes Delta Integral

Theorem \(\PageIndex{19}\)

Let \(g\) be a strictly increasing function on \(I\), \(f_1, f_2\in \mathcal{R}(g, I)\) and \(f_1\leq f_2\) on \(I\). Then
\[\begin{equation}
\label{39} \int\limits_a^b f_1(t)\Delta g(t)\leq \int\limits_a^b f_2(t)\Delta g(t).
\end{equation}\]

Proof

Since \(f_1, f_2\in \mathcal{R}(g, I)\), it follows that \(f_2-f_1\in \mathcal{R}(g, I)\). Moreover, we have
\[
f_2(t)-f_1(t)\geq 0,\quad t\in I.
\]
Then, we obtain
\[
\int\limits_a^b (f_2(t)-f_1(t))\Delta g(t)\geq 0.
\]
Now, we find
\[\begin{eqnarray*}
0&\leq& \int\limits_a^b (f_2(t)-f_1(t))\Delta g(t)\\ \\
&=& \int\limits_a^b f_2(t)\Delta g(t)-\int\limits_a^b f_1(t)\Delta g(t),
\end{eqnarray*}\]
whereupon we get the inequality \eqref{39}. This completes the proof.

Theorem \(\PageIndex{20}\)

Let \(g\) be a strictly increasing function on \(I\) and \(f\in \mathcal{R}(g, I)\). Then \(|f|\in \mathcal{R}(g, I)\) and we have the "triangle inequality" for Riemann-Stieltjes delta integrals
\[
\left|\int\limits_a^b f(t)\Delta g(t)\right|\leq \int\limits_a^b |f(t)|\Delta g(t).
\]

Proof

Fix \(\epsilon>0\) arbitrarily. Since \(f\in \mathcal{R}(g, I)\), there is a partition \(P=\{t_0, t_1, \ldots, t_n\}\) of \(I\) such that
\[
U(P, f, g)-L(P, f, g)<\epsilon.
\]
Define
\[\begin{eqnarray*}
m_j&=& \inf\limits_{t\in I_j} f(t),\\ \\
M_j&=& \sup\limits_{t\in I_j}f(t),\\ \\
m_j^\prime &=& \inf\limits_{t\in I_j} |f(t)|,\\ \\
M_j^\prime&=& \sup\limits_{t\in I_j} |f(t)|,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Then
\[\begin{eqnarray*}
m_j&\leq &f(t)\leq M_j,\\ \\
-m_j&\geq& f(t)\geq -M_j,\quad t\in I_j,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Combining these two inequalities, we find
\[
m_j-M_j\leq f(t)-f(y)\leq M_j-m_j,\quad t, y\in I_j,\quad j\in \{1, \ldots, n\}.
\]
Hence,
\[
|f(t)-f(y)|\leq M_j-m_j,\quad t, y\in I_j,\quad j\in \{1, \ldots, n\}.
\]
Applying the reverse triangle inequality
\[\begin{eqnarray*}
|f(t)|-|f(y)|&\leq& |f(t)-f(y)|\\ \\
&\leq& M_j-m_j,\quad t, y\in I_j,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Thus, \(M_j-m_j\) is an upper bound of \(|f(t)|-|f(y)|\), \(t, y\in I_j\), \(j\in \{1, \ldots, n\}\). On the other hand, we have
\[\begin{eqnarray*}
\sup\limits_{t, y\in I_j}\left(|f(t)|-|f(y)|\right)&=& \sup\limits_{t\in I_j}|f(t)|-\inf\limits_{y\in I_j}|f(y)|\\ \\
&=& M_j^\prime-m_j^\prime,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Hence, we have the inequality
\[
M_j^\prime-m_j^\prime\leq M_j-m_j,\quad j\in \{1, \ldots, n\}.
\]
For the difference of the upper and lower Darboux-Stieltjes sums of \[|f|\] with respect to the partition \[P\], we get
\[\begin{eqnarray*}
U(P, |f|, g)-L(P, |f|, g)&=& \sum\limits_{j=1}^n (M_j^\prime-m_j^\prime)\Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n (M_j-m_j)\Delta g_j\\ \\
&=& U(P, f, g)-L(P, f, g)\\ \\
&<& \epsilon,
\end{eqnarray*}\]
which is the \(\epsilon\)-criterion. Thus, \(|f|\in \mathcal{R}(g, I)\). Now, we will prove the "triangle inequality" for Riemann-Stieltjes delta integrals. Note that
\[
-|f|\leq f\leq |f|\quad \mbox{on}\quad I.
\]
Now, we find
\[
-\int\limits_a^b |f(t)|\Delta g(t)\leq \int\limits_a^b f(t)\Delta g(t)\leq \int\limits_a^b |f(t)|\Delta g(t),
\]
whereupon
\[
\left|\int\limits_a^b f(t)\Delta g(t)\right|\leq \int\limits_a^b |f(t)|\Delta g(t).
\]
This completes the proof.

Theorem \(\PageIndex{21}\)

Let \(g\) be a strictly increasing function on \(I\), \(f\in \mathcal{R}(g, I)\) and
\[
m\leq f(t)\leq M,\quad t\in I.
\]
Then
\[
m(g(b)-g(a))\leq \int\limits_a^b f(t)\Delta g(t)\leq M(g(b)-g(a)).
\]

Proof

We have
\[\begin{eqnarray*}
m(g(b)-g(a))&\leq& m\sum\limits_{j=1}^n \Delta g_j\\ \\
&\leq& L(P, f, g)\\ \\
&\leq& U(P, f, g)\\ \\
&\leq& M\sum\limits_{j=1}^n \Delta g_j\\ \\
&=& M(g(b)-g(a)).
\end{eqnarray*}\]
Using that \(f\in \mathcal{R}(g, I)\), we get
\[\begin{eqnarray*}
m(g(b)-g(a))&\leq& \sup\limits_{P\in \mathcal{P}(I)}L(P, f, g)\\ \\
&=& \int\limits_a^b f(t)\Delta g(t)\\ \\
&\leq& \inf\limits_{P\in \mathcal{P}(I)} U(P, f, g)\\ \\
&\leq&M(g(b)-g(a)).
\end{eqnarray*}\]
This completes the proof.

Theorem \(\PageIndex{22}\)

Let \(g\) be a strictly increasing function on \(I\), \(g^\Delta\) exists and is continuous on \(I\), and \(f\) is a bounded function on \(I\). Then \(f\in \mathcal{R}(g, I)\) if and only if \(fg^\Delta \in \mathcal{R}(E, I)\). Moreover,
\[\begin{equation}
\label{47}
\int\limits_a^b f(t)\Delta g(t)= \int\limits_a^b f(t)g^\Delta(t)\Delta t.
\end{equation}\]
Here \(E\) is the identity function, i.e., \(E(t)=t\), \(t\in I\).

Proof

We will prove the implications separately.

1. Let \(f\in \mathcal{R}(g, I)\). Because \(g^\Delta\) is continuous on \(I\), it is bounded on \(I\) and Riemann delta integrable on \(I\). Since \(f\) is a bounded function on \(I\), there is a positive constant \(K\) such that
\[
|f(t)|\leq K,\quad t\in I.
\]
Fix \(\epsilon>0\). Since \(f\in \mathcal{R}(g, I)\), there is a partition \(P_1\in \mathcal{P}(I)\) such that
\[\begin{equation}
\label{43} U(P_1, f, g)-L(P_1, f, g)<\frac{\epsilon}{2}.
\end{equation}\]
Because \(g^\Delta\) is Riemann delta integrable on \(I\), there is a partition \(P_2\in \mathcal{P}(I)\) so that
\[\begin{equation}
\label{44} U(P_2, g^\Delta, E)-L(P_2, g^\Delta, E)<\frac{\epsilon}{2K}.
\end{equation}\]
Let \(P=P_1\cup P_2=\{t_0, t_1, \ldots, t_n\}\). Then \(P\) is a refinement of \(P_1\) and \(P_2\). Then, we have the following inequalities
\[\begin{eqnarray*}
U(P, f, g)&\leq& U(P_1, f, g),\\ \\
L(P, f, g)&\geq& L(P_1, f, g),\\ \\
U(P, g^\Delta, E)&\leq& U(P_2, g^\Delta, E),\\ \\
L(P, g^\Delta, E)&\geq& L(P_2, g^\Delta, E).
\end{eqnarray*}\]
Hence, using \eqref{43} and \eqref{44}, we find
\[\begin{equation}
\label{45}
\begin{array}{lll}
U(P, f, g)-L(P, f, g)&\leq& U(P_1, f, g)-L(P_1, f, g)\\ \\
&<& \frac{\epsilon}{2}
\end{array}
\end{equation}\]
and
\[\begin{equation}
\label{46}
\begin{array}{lll}
U(P, g^\Delta, E)-L(P, g^\Delta, E)&\leq& U(P_2, g^\Delta, E)-L(P_2, g^\Delta, E)\\ \\
&<& \frac{\epsilon}{2K},
\end{array}
\end{equation}\]
respectively. Introduce the quantities
\[\begin{eqnarray*}
M_j^f&=& \sup\limits_{t\in I_j}f(t),\\ \\
M_j^{g^\Delta}&=& \sup\limits_{t\in I_j}g^\Delta(t),\\ \\
MP_j^{fg^\Delta}&=& \sup\limits_{t\in I_j}(fg^\Delta)(t),\\ \\
m_j^f&=& \inf\limits_{t\in I_j}f(t),\\ \\
m_j^{g^\Delta}&=& \inf\limits_{t\in I_j}g^\Delta(t),\\ \\
m_j^{fg^\Delta}&=& \inf\limits_{t\in I_j}(fg^\Delta)(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
By the delta mean value theorem, it follows that for each \(j\in \{1, \ldots, n\}\) there are \(\xi_j, \tau_j\in [t_{j-1}, t_j]\) such that
\[
g^\Delta(\tau_j)\Delta t_j\leq \Delta g_j\leq g^\Delta(\xi_j)\Delta t_j.
\]
Then, for any \(j\in \{1, \ldots, n\}\) and any \(y\in I_j\), one has
\[\begin{eqnarray*}
(fg^\Delta)(\tau_j)-(fg^\Delta)(y)&=& f(\tau_j)g^\Delta(\tau_j)-f(y)g^\Delta(y)\\ \\
&=& f(\tau_j)g^\Delta(\tau_j)-f(y)g^\Delta(\tau_j)+f(y)g^\Delta(\tau_j)-f(y)g^\Delta(y)\\ \\
&=& (f(\tau_j)-f(y))g^\Delta(\tau_j)+f(y)(g^\Delta(\tau_j)-g^\Delta(y))\\ \\
&\leq& |f(\tau_j)-f(y)|g^\Delta(\tau_j)+|f(y)||g^\Delta(\tau_j)-g^\Delta(y)|\\ \\
&\leq& (M_j^f-m_j^f)\frac{\Delta g_j}{\Delta t_j}+K\left(M_j^{g^\Delta}-m_j^{g^\Delta}\right),
\end{eqnarray*}\]
which is an upper bound of \((fg^\Delta)(t)-(fg^\Delta)(z)\) for \(t, z\in I_j\), \(j\in \{1, \ldots, n\}\). On the other hand, we have
\[\begin{eqnarray*}
\sup\limits_{t, z\in I_j}\left((fg^\Delta)(t)-(fg^\Delta)(z)\right)&=& \sup\limits_{t\in I_j}(fg^\Delta)(t)+\sup\limits_{z\in I_j}\left(-(fg^\Delta)(z)\right)\\ \\
&=& M_j^{fg^\Delta}-\inf\limits_{z\in I_j}(fg^\Delta)(z)\\ \\
&=& M_j^{fg^\Delta}-m_j^{fg^\Delta},\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Consequently
\[
M_j^{fg^\Delta}-m_j^{fg^\Delta}\leq (M_j^f-m_j^f)\frac{\Delta g_j}{\Delta t_j}+K\left(M_j^{g^\Delta}-m_j^{g^\Delta}\right),\quad j\in \{1, \ldots, n\}.
\]
Hence, applying \eqref{45} and \eqref{46}, we arrive at
\[\begin{eqnarray*}
U(P, fg^\Delta, E)-L(P, fg^\Delta, E)&=& \sum\limits_{j=1}^n \left(M_j^{fg^\Delta}-m_j^{fg^\Delta}\right)\Delta t_j
\end{eqnarray*}\]
\[\begin{eqnarray*}
&\leq& \sum\limits_{j=1}^n \left((M_j^f-m_j^f)\frac{\Delta g_j}{\Delta t_j}+K\left(M_j^{g^\Delta}-m_j^{g^\Delta}\right)\right)\Delta t_j\\ \\
&=& \sum\limits_{j=1}^n(M_j^f-m_j^f)\Delta g_j+K\sum\limits_{j=1}^n\left(M_j^{g^\Delta}-m_j^{g^\Delta}\right)\Delta t_j\\ \\
&=& U(P, f, g)-L(P, f, g)+K\left(U(P, g^\Delta, E)-L(P, g^\Delta, E)\right)\\ \\
&<& \frac{\epsilon}{2}+K\frac{\epsilon}{2K}\\ \\
&=& \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ \\
&=& \epsilon,
\end{eqnarray*}\]
which is the \(\epsilon\)-criterion. This implies that \(fg^\Delta\in \mathcal{R}(E, I)\). Now, we will prove that \eqref{47} holds. We have
\[\begin{eqnarray*}
f(\tau_j)g^\Delta(\tau_j)&\leq& M_j^fg^\Delta(\tau_j)\\ \\
&\leq& M_j^f \frac{\Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Thus, \(M_j^f \frac{\Delta g_j}{\Delta t_j}\) is an upper bound of \(f(t)g^\Delta(t)\), \(t\in I_j\), \(j\in \{1, \ldots, n\}\). Therefore
\[
M_j^{fg^\Delta}\leq M_j^f \frac{\Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\}.
\]
Hence,
\[
U(P, fg^\Delta, E)\leq U(P, f, g).
\]
Taking infimum of both sides of the last inequality over \(P\in \mathcal{P}(I)\), we get
\[\begin{equation}
\label{43.1} \int\limits_a^b f(t)g^\Delta(t)\Delta t\leq \int\limits_a^b f(t)\Delta g(t).
\end{equation}\]
On the other hand, we have
\[
f(\xi_j)g^\Delta(\xi_j)\geq m_j^f \frac{\Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\}.
\]
Thus, \(m_j^f \frac{\Delta g_j}{\Delta t_j}\), \(j\in \{1, \ldots, n\}\), is a lower bound of \(f(t)g^\Delta(t)\), \(t\in I_j\), \(j\in \{1, \ldots, n\}\). Therefore
\[
m_j^{fg^\Delta}\geq m_j^f \frac{\Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\},
\]
and
\[
L(P, fg^\Delta, E)\geq L(P, f, g).
\]
Taking supremum of both sides of the last inequality over \(P\in \mathcal{P}(I)\), we get
\[
\int\limits_a^b f(t)g^\Delta(t)\Delta t\geq \int\limits_a^b f(t)\Delta g(t).
\]
By the last inequality and the inequality \eqref{43.1}, we obtain the equation \eqref{47}.

2. Let \(fg^\Delta \in \mathcal{R}(E, I)\). We will use the notations in the previous point. Fix \(\epsilon>0\) arbitrarily. Then, there exists a partition \(P=\{t_0, t_1, \ldots, t_n\}\) of the interval \(I\) so that
\[\begin{equation}
\label{48} U(P, fg^\Delta, E)-L(P, fg^\Delta, E)<\frac{\epsilon}{2}.
\end{equation}\]
Note that
\[\begin{eqnarray*}
M_j^{fg^\Delta}&\geq& f(\xi_j)g^\Delta (\xi_j)\\ \\
&\geq& f(\xi_j)\frac{\Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\},
\end{eqnarray*}\]
whereupon \(M_j^{fg^\Delta}\) is an upper bound of \(f(t)\frac{\Delta g_j}{\Delta t_j}\), \(t\in I_j\), \(j\in \{1, \ldots, n\}\). On the other hand, we have
\[
\sup\limits_{t\in I_j}\left(f(t)\frac{\Delta g_j}{\Delta t_j}\right)=\frac{M_j^f \Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\}.
\]
Consequently
\[
M_j^{fg^\Delta}\geq \frac{M_j^f \Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\},
\]
and
\[\begin{equation}
\label{49}
M_j^f \Delta g_j\leq M_j^{fg^\Delta}\Delta t_j,\quad j\in \{1, \ldots, n\}.
\end{equation}\]
Next, we have
\[\begin{eqnarray*}
m_j^{fg^\Delta}&\leq& f(\tau_j)g^\Delta(\tau_j)\\ \\
&\leq& \frac{f(\tau_j)\Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
So, \(m_j^{fg^\Delta}\) is a lower bound of \(\frac{f(t)\Delta g_j}{\Delta t_j}\), \(t\in I_j\), \(j\in \{1, \ldots, n\}\). On the other hand,
\[
\inf\limits_{t\in I_j}\frac{f(t)\Delta g_j}{\Delta t_j}=\frac{m_j^f\Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\}.
\]
Therefore
\[
m_j^{fg^\Delta}\leq \frac{m_j^f \Delta g_j}{\Delta t_j},\quad j\in \{1, \ldots, n\},
\]
and
\[\begin{equation}
\label{50}
m_j^{fg^\Delta}\Delta t_j\leq m_j^f \Delta g_j,\quad j\in \{1, \ldots, n\}.
\end{equation}\]
Now, applying \eqref{48}, \eqref{49} and \eqref{50}, we find
\[\begin{eqnarray*}
U(P, f, g)-L(P, f, g)&=& \sum\limits_{j=1}^n M_j^f \Delta g_j-\sum\limits_{j=1}^n m_j^f \Delta g_j\\ \\
&\leq& \sum\limits_{j=1}^n M_j^{fg^\Delta}\Delta t_j-\sum\limits_{j=1}^n m_j^{fg^\Delta}\Delta t_j\\ \\
&=& U(P, fg^\Delta, E)-L(P, fg^\Delta, E)\\ \\
&<& \epsilon,
\end{eqnarray*}\]
which is the \(\epsilon\)-criterion. Therefore \(f\in \mathcal{R}(g, I)\). By \eqref{49}, we get
\[
U(P, f, g)\leq U(P, fg^\Delta, E).
\]
We take infimum of both sides of the last inequality over \(P\in \mathcal{P}(I)\) and we find
\[\begin{equation}
\label{52} \int\limits_a^b f(t)\Delta g(t)\leq \int\limits_a^b f(t)g^\Delta(t)\Delta t.
\end{equation}\]
By \eqref{50}, we find
\[
L(P, f, g)\geq L(P, fg^\Delta, E).
\]
We take supremum of both sides of the last inequality over \(P\in \mathcal{P}(I)\) and we obtain
\[
\int\limits_a^b f(t)\Delta g(t)\geq \int\limits_a^g f(t)g^\Delta(t)\Delta t.
\]
By the last inequality and the inequality \eqref{52}, we get \eqref{47}. This completes the proof.

Theorem \(\PageIndex{23}\)(Delta Integration by Parts

Let \(g\) be a strictly increasing function on \(I\) such that \(g^\Delta\) exists and is continuous on \(I\). Let also, \(f\) be a strictly increasing bounded function on \(I\) such that \(f^\Delta\) exists on \(I\) and is continuous on \(I\). Then
\[\begin{equation}
\label{58} \int\limits_a^b f(t)\Delta g(t)=f(t)g(t)\bigg|_{t=a}^{t=b}-\int\limits_a^b g^\sigma(t) \Delta f(t).
\end{equation}\]

Proof

Since \(f^\Delta\) exists and is continuous on \(I\), we have that \(f\) is Riemann delta integrable on \(I\). Because \(g^\Delta\) is continuous on \(I\), it is Riemann delta integrable on \(I\). From here, we get that \(fg^\Delta\) is Riemann delta integrable on \(I\), i.e., \(fg^\Delta\in \mathcal{R}(E, I)\). Now, applying Theorem \(\PageIndex{22}\), we find that \(f\) is Riemann-Stieltjes delta integrable on \(I\) and
\[\begin{equation}
\label{56} \int\limits_a^b f(t)\Delta g(t)= \int\limits_a^b f(t)g^\Delta (t)\Delta t.
\end{equation}\]
Now, delta integration by parts gives
\[\begin{equation}
\label{57} \int\limits_a^b f(t)g^\Delta(t)\Delta t=f(t)g(t)\bigg|_{t=a}^{t=b}-\int\limits_a^b g^\sigma(t) f^\Delta(t)\Delta t.
\end{equation}\]
Since \(g^\sigma f^\Delta\in \mathcal{R}(E, I)\) and \(f\) is a strictly increasing function on \(I\), applying Theorem \(\PageIndex{22}\), we get \(g^\sigma\in \mathcal{R}(f, I)\) and
\[
\int\limits_a^b g^\sigma(t) f^\Delta(t)\Delta t=\int\limits_a^b g^\sigma(t)\Delta f(t).
\]
By the last relation and the relations \eqref{56}, \eqref{57}, we arrive at \eqref{58}. This completes the proof.

Theorem \(\PageIndex{24}\)

Let \(f\in \mathcal{C}(I)\), \(g\) be a strictly increasing function on \(I\) and \(f\in \mathcal{R}(g, I)\). Then, we have the following estimate
\[
\left|\int\limits_a^b f(t)\Delta g(t)\right|\leq \max\limits_{t\in I} |f(t)|(g(b)-g(a)).
\]

Proof

Since \(f\in \mathcal{C}(I)\) and \(I\) is a compact set, there exists \(\max\limits_{t\in I}|f(t)\). Applying the "triangle inequality" for Riemann-Stieltjes delta integrals, Theorem \(\PageIndex{20}\), we obtain
\[\begin{equation}
\label{60} \left|\int\limits_a^b f(t)\Delta g(t)\right|\leq \int\limits_a^b |f(t)|\Delta g(t).
\end{equation}\]
Now, we find
\[\begin{equation}
\label{61} \int\limits_a^b |f(t)|\Delta g(t)\leq \max\limits_{t\in I} |f(t)|\int\limits_a^b 1 \Delta g(t).
\end{equation}\]
Next, we find
\[
\int\limits_a^b 1\Delta g(t)=g(b)-g(a).
\]
By the last equation and the equations \eqref{60}, \eqref{61}, we arrive at
\[\begin{eqnarray*}
\left|\int\limits_a^b f(t)\Delta g(t)\right|&\leq& \int\limits_a^b |f(t)|\Delta g(t)\\ \\
&\leq& \max\limits_{t\in I} |f(t)|\int\limits_a^b 1 \Delta g(t)\\ \\
&=& \max\limits_{t\in [a, b]}|f(t)| (g(b)-g(a)).
\end{eqnarray*}\]
This completes the proof.

Theorem \(\PageIndex{1}\)

Let \(\{f_n\}_{n\in \mathbb{N}}\) be a sequence of continuous functions on \(I\) that converges uniformly to the continuous function \(f\) on \(I\). Let also, \(g\) be a strictly increasing function on \(I\) and \(f_n, f\in \mathcal{R}(g, I)\), \(n\in \mathbb{N}\). Then
\[
\lim\limits_{n\to \infty} \int\limits_a^b f_n(t)\Delta g(t)=\int\limits_a^b f(t)\Delta g(t).
\]

Proof

Applying Theorem \(\PageIndex{24}\), we get
\[\begin{eqnarray*}
\left|\int\limits_a^b f_n(t)\Delta g(t)-\int\limits_a^b f(t)\Delta g(t)\right|&=& \left|\int\limits_a^b (f_n(t)-f(t))\Delta g(t)\right|\\ \\
&\leq& \max\limits_{t\in [a, b]}|f_n(t)-f(t)|(g(b)-g(a))\\ \\
&\to& 0,\quad \mbox{as}\quad n\to \infty.
\end{eqnarray*}\]
This completes the proof.