4: Properties of the Riemann-Stieltjes Delta Integral
- Page ID
- 207602
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Now, that we have clarified the definitions of Riemann-Stieltjes and Darboux-Stieltjes delta integrals, we move on to study their properties. Since the construction for Darboux-Stieltjes delta integral is more instructive, we shall prove these results using the Darboux-Stieltjes definition.
We start with the following preliminary result.
Let \(g\) be a strictly increasing function on \(I\). For any two bounded functions \(f_1, f_2: I\to \mathbb{R}\), we have
\[\begin{equation}
\label{23} \overline{\int\limits_a^b} (f_1+f_2)(t)\Delta g(t)\leq \overline{\int\limits_a^b} f_1(t)\Delta g(t)+\overline{\int\limits_a^b} f_2(t)\Delta g(t)
\end{equation}\]
and
\[\begin{equation}
\label{24}\underline{\int\limits_a^b }(f_1+f_2)T)\Delta g(t)\geq \underline{\int\limits_a^b} f_1(t)\Delta g(t)+\underline{\int\limits_a^b}f_2(t)\Delta g(t).
\end{equation}\]
Proof
Take \(P=\{t_0, t_1, \ldots, t_n\}\) and
\[\begin{eqnarray*}
M_j^{f_1}&=& \sup\limits_{t\in I_j}f_1(t),\\ \\
M_j^{f_2}&=& \sup\limits_{t\in I_j}f_2(t),\\ \\
M_j^{f_1+f_2}&=& \sup\limits_{t\in I_j}(f_1+f_2)(t),\\ \\
m_j^{f_1}&=& \inf\limits_{t\in I_j}f_1(t),\\ \\
m_j^{f_2}&=& \inf\limits_{t\in I_j}f_2(t),\\ \\
m_j^{f_1+f_2}&=& \inf\limits_{t\in I_j}(f_1+f_2)(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Note that for any \(t\in I_j\), \(j\in \{1, \ldots, n\}\), one has
\[
f_1(t)+f_2(t)_\leq M_j^{f_1}+M_j^{f_2},\quad j\in \{1, \ldots, n\},
\]
and
\[
f_1(t)+f_2(t)\geq m_j^{f_1}+m_j^{Pf_2},\quad j\in \{1, \ldots, n\}.
\]
By taking supremum in the first inequality over \(t\in I_j\), \(j\in \{1, \ldots, n\}\), and infimum in the second inequality over \(t\in I_j\), \(j\in \{1, \ldots, n\}\), we get
\[
M_j^{f_1+f_2}\leq M_j^{f_1}+M_j^{f_2},\quad j\in \{1, \ldots, n\},
\]
and
\[
m_j^{f_1+f_2}\geq m_j^{f_1}+m_j^{f_2},\quad j\in \{1, \ldots, n\},
\]
respectively. Hence,
\[
U(P, f_1+f_2, g)\leq U(P, f_1, g)+U(P_2, f, g)
\]
and
\[
L(P, f_1+f_2, g)\geq L(P, f_1, g)+L(P, f_2, g)
\]
for any partition \(P\in \mathcal{P}(I)\). By taking infimum in the first inequality over \(P\in \mathcal{P}(I)\) and supremum in the second inequality over \(P\in \mathcal{P}(I)\), we obtain \eqref{23} and \eqref{24}, respectively. This completes the proof.
Let \(g\) be a strictly increasing function on \(I\). Let also, \(f_1, f_2\in \mathcal{R}(g, I)\). Then, for any \(\alpha, \beta\in \mathbb{R}\), one has \(\alpha f_1+\beta f_2\in \mathcal{R}(g, I)\).
Proof
Firstly, we will prove the assertion for $\alpha=\beta=1$. By Lemma \(\PageIndex{12}\), we get
\[\begin{eqnarray*}
\overline{\int\limits_a^b }(f_1+f_2)(t)\Delta g(t)&\leq& \overline{\int\limits_a^b} f_1(t)\Delta g(t)+\overline{\int\limits_a^b}f_2(\Delta g(t)\\ \\
&=& \int\limits_a^b f_1(t)\Delta g(t)+\int\limits_a^b f_2(t)\Delta g(t)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\underline{\int\limits_a^b}(f_1+f_2)(t)\Delta g(t)&\geq& \underline{\int\limits_a^b} f_1(t)\Delta g(t)+\underline{\int\limits_a^b}f_2(t)\Delta g(t)\\ \\
&=& \int\limits_a^b f_1(t)\Delta g(t)+\int\limits_a^b f_2(t)\Delta g(t).
\end{eqnarray*}\]
Therefore
\[\begin{eqnarray*}
\underline{\int\limits_a^b}(f_1+f_2)(t)\Delta g(t)&=& \overline{\int\limits_a^b}(f_1+f_2)(t)\Delta g(t)\\ \\
&=& \int\limits_a^b f_1(t)\Delta g(t)+\int\limits_a^b f_2(t)\Delta g(t)
\end{eqnarray*}\]
and \(f_1+f_2\in \mathcal{R}(g, I)\).
Now, we will prove that \(\alpha f_1\in \mathcal{R}(g, I)\) for any \(\alpha\in \mathbb{R}\). If \(\alpha=0\), then the assertion is evident. Let \(\alpha>0\) and \(P=\{t_0, t_1, \ldots, t_n\}\in \mathcal{P}(I)\). Let also,
\[\begin{eqnarray*}
M_j&=& \sup\limits_{t\in I_j}f(t),\\ \\
m_j&=& \inf\limits_{t\in I_j} f(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Then
\[\begin{eqnarray*}
\alpha M_j&=& \alpha\sup\limits_{t\in I_j}f_1(t)\\ \\
&=& \sup\limits_{t\in I_j} (\alpha f_1)(t),\quad j\in \{1, \ldots, n\},
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\alpha m_j&=& \alpha \inf\limits_{t\in I_j} f_1(t)\\ \\
&=& \inf\limits_{t\in I_j}(\alpha f_1)(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Hence, we conclude that
\[
\alpha L(P, f_1, g)=L(P, \alpha f_1, g)
\]
and
\[
\alpha U(P, f_1, g)=U(P, \alpha f_1, g).
\]
Since \(f_1\in \mathcal{R}(g, I)\), we have that
\[
L(P, f_1, g)= U(P, f_1, g)
\]
and
\[
\alpha L(P, f_1, g)=\alpha U(P, f_1, g).
\]
Therefore
\[
U(P, \alpha f_1, g)=L(P, \alpha f_1, g)
\]
for any partition \(P\in \mathcal{P}(I)\). Therefore \(\alpha f_1\in \mathcal{R}_1(g, I)\) and
\[
\int\limits_a^b (\alpha f_1)(t)\Delta g(t)=\alpha \int\limits_a^b f_1(t)\Delta g(t).
\]
Let now, \(\alpha<0\). Then
\[
\alpha M_j=\inf\limits_{t\in I_j}(\alpha f_1)(t),\quad j\in \{1, \ldots, n\},
\]
and
\[
\alpha m_j=\sup\limits_{t\in I_j} (\alpha f_1)(t),\quad j\in \{1, \ldots, n\}.
\]
From here, we conclude that
\[\begin{eqnarray*}
\alpha U(P, f_1, g)&=& L(P, \alpha f_1, g),\\ \\
\alpha L(P, f_1, g)&=& U(P, \alpha f_1, g).
\end{eqnarray*}\]
Consequently
\[
U(P, \alpha f_1, g)=L(P, \alpha f_1, g)
\]
for any \(P\in \mathcal{P}(I)\) and \(\alpha f_1\in \mathcal{R}(g, I)\), and
\[
\int\limits_a^b (\alpha f_1)(t)\Delta g(t)=\alpha \int\limits_a^b f_1(t)\Delta g(t).
\]
As above, one has \(\beta f_2\in \mathcal{R}(g, I)\) for any \(\beta\in \mathbb{R}\). Then \(\alpha f_1+\beta f_2\in \mathcal{R}(g, I)\) and
\[
\int\limits_a^b (\alpha f_1+\beta f_2)(t)\Delta g(t)=\alpha \int\limits_a^b f_1(t)\Delta g(t)+\beta \int\limits_a^b f_2(t)\Delta g(t).
\]
This completes the proof.
Let \(g\) be a strictly increasing function on \(I\) Let also, \(f\in \mathcal{R}(g, I)\). Then \(f^2\in \mathcal{R}(g, I)\).
Proof
Since \(f\in \mathcal{R}(g, I)\), then \(f\) is a bounded function on \(I\). Thus, there exists a constant \(K>0\) such that
\[
|f(t)|\leq K,\quad t\in I.
\]
Take \(\epsilon>0\) arbitrarily. Then, applying the $\epsilon$-criterion for Darboux -Stieltjes delta integrability, we get that there is a \(P\in \mathcal{P}(I)\) such that
\[
U(P, f, g)-L(P, f, g)<\frac{\epsilon}{2K}.
\]
Let
\[\begin{eqnarray*}
M_j&=&\sup\limits_{t\in I_j} f(t),\\ \\
m_j&=& \inf\limits_{t\in I_j} f(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Note that for any \(t, y\in I_j\), \(j\in \{1, \ldots, n\}\), one has
\[\begin{eqnarray*}
(f(t))^2-(f(y))^2&=& (f(t)-f(y))(f(t)+f(y))\\ \\
&\leq& (f(t)+f(y))|f(t)-f(y)|\\ \\
&\leq& 2K |f(t)-f(y)|\\ \\
&\leq& 2K(M_j-m_j),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Since \(t\) and \(y\) are arbitrary in \(I\), we conclude that \(2K(M_j-m_j)\) is an upper bound for the difference \((f(t))^2-(f(y))^2\) for any \(t, y\in I_j\), \(j\in \{1, \ldots, n\}\). On the other hand, the supremum of this quantity over all pairs \(t, y\in I_j\), \(j\in \{1, \ldots, n\}\), is
\[\begin{eqnarray*}
\sup\limits_{t, y\in I_j}\left((f(t))^2-(f(y))^2\right)&=& \sup\limits_{t\in I_j}(f(t))^2+\sup\limits_{y\in I_j}\left(-(f(y))^2\right)\\ \\
&=& \sup\limits_{t\in I_j}(f(t))^2-\inf\limits_{t\in I_j}(f(y))^2\\ \\
&\leq& M_j^{f^2}-m_j^{f^2},\quad j\in \{1, \ldots, n\},
\end{eqnarray*}\]
where
\[\begin{eqnarray*}
M_j^{f^2}&=& \sup\limits_{t\in I_j}(f(t))^2,\\ \\
m_j^2&=& \inf\limits_{t\in I_j}(f(t))^2,\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Therefore
\[
M_j^{f^2}-m_j^{f^2}\leq 2K (M_j-m_j),\quad j\in \{1, \ldots, n\}.
\]
Thus, by constructing the Darboux-Stieltjes sum, for the function \(f^2\) with respect to the partition \(P\), we get
\[\begin{eqnarray*}
U(P, f^2, g)-L(P, f^2, g)&=& \sum\limits_{j=1}^n \left(M_j^{f^2}-m_j^{f^2}\right)\Delta g_j\\ \\
&\leq& 2K\sum\limits_{j=1}^n (M_j-m_j)\Delta g_j\\ \\
&=& 2K(U(P, f, g)-L(P, f, g))\\ \\
&<& 2K\frac{\epsilon}{2K}\\ \\
&=& \epsilon,
\end{eqnarray*}\]
which is the \(\epsilon\)-criterion for Darboux-Stieltjes delta integrability. This proves \(f^2\in \mathcal{R}(g, I)\).
Let \(f_1, f_2\in \mathcal{R}(g, I)\) and \(g\) be a strictly increasing function on \(I\). Then \(f_1f_2\in \mathcal{R}(g, I)\).
Proof
Observe that
\[
f_1f_2=\frac{1}{4}(f_1+f_2)^2-\frac{1}{4}(f_1-f_2)^2\quad \mbox{on}\quad I.
\]
By Theorem \(\PageIndex{13}\}, we have that
\[f_1+f_2, f_1-f_2\in \mathcal{R}(g, I).\]
Hence, applying Theorem \(\PageIndex{14}\}, we find that
\[(f_1+f_2)^2, (f_1-f_2)^2\in \mathcal{R}(g, I).\]
Again applying Theorem \(\PageIndex{13}\), we obtain that
\[\frac{1}{4}(f_1+f_2)^2-\frac{1}{4}( f_1-f_2)^2\in \mathcal{R}(g, I).\]
i.e., \(f_1f_2\in \mathcal{R}(g, I)\). This completes the proof.
Let \(f\in \mathcal{R}(g, I)\), \(g\) be a strictly increasing function on \(I\) and \(c\in I\). Then \(f\in \mathcal{R}(g, [a, c])\) and \(f\in \mathcal{R}(g, [c, b])\), and
\[\begin{equation}
\label{29}\int\limits_a^b f(t)\Delta g(t)=\int\limits_a^c f(t)\Delta g(t)+\int\limits_c^b f(t)\Delta g(t).
\end{equation}\]
Proof
Take \(\epsilon>0\) arbitrarily. Since \(f\in \mathcal{R}(g, I)\), by the \(\epsilon\)-criterion for Darboux-Stieltjes delta integrability, it follows that there is a partition \(P\in \mathcal{P}(I)\) such that
\[
U(P, f, g)-L(P, f, g)<\epsilon.\]
Define \(P^\prime =P\cup \{c\}\) and
\[\begin{eqnarray*}
P_1&=& P^\prime \cap [a, c],\\ \\
P_2&=& P^\prime \cap [c, b].
\end{eqnarray*}\]
Thus, by definition, we have
\[\begin{eqnarray*}
U(P_1, f, g)+U(P_2, f, g)&=& U(P^\prime, f, g)\\ \\
&\leq& \overline{\int\limits_a^b}f(t)\Delta g(t)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
L(P_1, f, g)+L(P_2, f, g)&=& L(P^\prime, f, g)\\ \\
&\geq& \underline{\int\limits_a^b} f(t)\Delta g(t).
\end{eqnarray*}\]
Hence,
\[\begin{eqnarray*}
\epsilon&>& U(P, f, g)-L(P, f, g)\\ \\
&\geq& \overline{\int\limits_a^b} f(t)\Delta g(t)-\underline{\int\limits_a^b} f(t)\Delta g(t)\\ \\
&\geq& U(P_1, f, g)+U(P_2, f, g)-L(P_1, f, g)-L(P_2, f, g)\\ \\
&=& \left(U(P_1, f, g)-L(P_1, f, g)\right)+\left(U(P_2, f, g)-L(P_2, f, g)\right).
\end{eqnarray*}\]
Since
\[
U(P_1, f, g)-L(P_1, f, g)\geq 0
\]
and
\[
U(P_2, f, g)-L(P_2, f, g)\geq 0,
\]
we get that
\[
U(P_1, f, g)-L(P_1, f, g)<\epsilon
\]
and
\[
U(P_2, f, g)-L(P_2, f, g)<\epsilon,
\]
which, by the \(\epsilon\)-criterion, imply that \(f\in \mathcal{R}(g, [a, c])\) and \(f\in \mathcal{R}(g, [c, b])\). Now, we will show \eqref{29}. Let \(Q=\{y_0, y_1, \ldots, y_n\}\) be an arbitrary partition of \(I\). Define \(Q^\prime=Q\cup \{c\}\) and
\[\begin{eqnarray*}
Q_1&=& Q^\prime \cap [a, c],\\ \\
Q_2&=& Q^\prime \cap [c, b].
\end{eqnarray*}\]
By the definition of the Darboux-Stieltjes delta integral, we have
\[\begin{eqnarray*}
\int\limits_a^c f(t)\Delta g(t)+\int\limits_c^b f(t)\Delta g(t)&\leq& U(Q_1, f, g)+U(Q_2, f, g)\\ \\
&=& U(Q^\prime, f, g)\\ \\
&\leq& U(Q, f, g)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\int\limits_a^c f(t)\Delta g(t)+\int\limits_c^b f(t)\Delta g(t)&\geq& U(Q_1, f, g)+U(Q_2, f, g)\\ \\
&=& L(Q^\prime, f, g)\\ \\
&\geq& L(Q, f, g),
\end{eqnarray*}\]
where we have used that \(Q^\prime\) is a refinement of \(Q\). Because \(Q\in \mathcal{P}(I)\) was arbitrarily chosen, we find
\[\begin{eqnarray*}
\int\limits_a^c f(t)\Delta g(t)+\int\limits_c^b f(t)\Delta g(t)&\leq& \inf\limits_{Q\in \mathcal{P}(I)}U(Q, f, g)\\ \\
&=& \int\limits_a^b f(t)\Delta g(t)
\end{eqnarray*}\]
and
\[\begin{eqnarray*}
\int\limits_a^c f(t)\Delta g(t)+\int\limits_c^b f(t)\Delta g(t)&\geq& \sup\limits_{Q\in \mathcal{P}(I)}L(Q, f, g)\\ \\
&=& \int\limits_a^b f(t)\Delta g(t)
\end{eqnarray*}\]
and hence, by sandwiching, we get the desired equality. This completes the proof.
Let \(g\) be a strictly increasing function on \(I\). For \(f\in \mathcal{R}(g, I)\), define
\[
\int\limits_b^a f(t)\Delta g(t)=-\int\limits_a^b f(t)\Delta g(t).
\]
This is motivated by Theorem \(\PageIndex{16}\). Namely, if we want to extend Theorem \(\PageIndex{16}\) to work for any \(c, d, e\in I\) in any order as well, we would have
\[
\int\limits_c^e f(t)\Delta g(t)=\int\limits_c^d f(t)\Delta g(t)+\int\limits_d^e f(t)\Delta g(t).\]
Since this is extended to be defined for \(c\), \(d\), \(e\) in any order, we can also set \(e=c\) to get
\[\begin{eqnarray*}
0&=& \int\limits_c^c f(t)\Delta g(t)\\ \\
&=& \int\limits_c^d f(t)\Delta g(t)+\int\limits_d^c f(t)\Delta g(t),
\end{eqnarray*}\]
whereupon
\[
\int\limits_c^d f(t)\Delta g(t)=-\int\limits_d^c f(t)\Delta g(t),
\]
agreeing with the above definition.
Let \(g_1\) and \(g_2\) be strictly increasing functions on \(I\) and \(f\in \mathcal{R}(g_1, I)\), \(f\in \mathcal{R}(g_2, I)\). Then \(f\in \mathcal{R}(g_1+g_2, I)\) and
\[\begin{equation}
\label{34} \int\limits_a^b f(t)\Delta (g_1+g_2)(t)=\int\limits_a^b f(t)\Delta g_1(t)+\int\limits_a^b f(t)\Delta g_2(t).
\end{equation}\]
Proof
Let \(\epsilon>0\) be arbitrarily chose. By the \(\epsilon\)-criterion, it follows that there exists a partition \(P=\{t_0, t_1, \ldots, t_n\}\) of \(I\) such that
\[
U(P, f, g_1)-L(P, f, g_1)<\frac{\epsilon}{2}\quad \mbox{and}\quad U(P, f, g_2)-L(P, f, g_2)<\frac{\epsilon}{2}.
\]
Note that
\[
\left(\Delta(g_1+g_2)\right)_j=\left(\Delta g_1\right)_j+\left(\Delta g_2\right)_j,\quad j\in \{1, \ldots, n\}.
\]
Let
\[\begin{eqnarray*}
M_j&=& \sup\limits_{t\in I_j}f(t),\\ \\
m_j&=& \inf\limits_{t\in I_j} f(t),\quad j\in \{1, \ldots, n\}.
\end{eqnarray*}\]
Then
\[\begin{equation}
\label{32}
\begin{array}{lll}
U(P, f, g_1+g_2)&=& \sum\limits_{j=1}^n M_j\left(\Delta(g_1+g_2)\right)_j\\ \\
&=& \sum\limits_{j=1}^n M_j\left((\Delta g_1)_j+(\Delta g_2)_j\right)\\ \\
&=& \sum\limits_{j-=1}^n M_j(\Delta g_1)_j+\sum\limits_{j=1}^n M_j(\Delta g_2)_j\\ \\
&=& U(P, f, g_1)+U(P, f, g_2)
\end{array}
\end{equation}\]
and
\[\begin{equation}
\label{33}
\begin{array}{lll}
L(P, f, g_1+g_2)&=& \sum\limits_{j=1}^n m_j\left(\Delta(g_1+g_2)\right)_j\\ \\
&=& \sum\limits_{j=1}^n m_j\left((\Delta g_1)_j+(\Delta g_2)_j\right)\\ \\
&=& \sum\limits_{j-=1}^n m_j(\Delta g_1)_j+\sum\limits_{j=1}^n m_j(\Delta g_2)_j\\ \\
&=& L(P, f, g_1)+L(P, f, g_2).
\end{array}
\end{equation}\]
Therefore
\[
U(p, f, g_1+g_2)+
L(P, f, g_1+g_2)= U(P, f, g_1)+U(P, f, g_2)-L(P, f, g_1)-L(P, f, g_2)
\]
\[\begin{eqnarray*}
&=& (U(P, f, g_1)-L(P, f, g_1))+(U(P, f, g_2)-L(P, f, g_2))\\ \\
&<& \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ \\
&=& \epsilon,
\end{eqnarray*}\]
which is the \(\epsilon\)-criterion for Darboux-Stieltjes delta integrability. Thus, \(f\in \mathcal{R}(g_1+g_2, I)\). Next, taking infimum in \eqref{32} over \(P\in \mathcal{P}(I)\), we arrive at
\[\begin{eqnarray*}
\overline{\int\limits_a^b}f(t)\Delta(g_1+g_2)(t)&\leq& \overline{\int\limits_a^b}f(t)\Delta g_1(t)+\overline{\int\limits_a^b}f(t)\Delta g_2(t)\\ \\
&=& \int\limits_a^b f(t)\Delta g_1(t)+\int\limits_a^b f(t)\Delta g_2(t)
\end{eqnarray*}\]
and taking supremum in \eqref{33} over \(P\in \mathcal{P}(I)\), we get
\[\begin{eqnarray*}
\underline{\int\limits_a^b}f(t)\Delta(g_1+g_2)(t)&\leq& \underline{\int\limits_a^b}f(t)\Delta g_1(t)+\underline{\int\limits_a^b}f(t)\Delta g_2(t)\\ \\
&=& \int\limits_a^b f(t)\Delta g_1(t)+\int\limits_a^b f(t)\Delta g_2(t).
\end{eqnarray*}\]
Therefore
\[\begin{eqnarray*}
\int\limits_a^b f(t)\Delta g_1(t)+\int\limits_a^b f(t)\Delta g_2(t)&\leq& \underline{\int\limits_a^b}f(t)\Delta(g_1+g_2)(t)\\ \\
&\leq& \overline{\int\limits_a^b}f(t)\Delta(g_1+g_2)(t)\\ \\
&\leq& \int\limits_a^b f(t)\Delta g_1(t)_+\int\limits_a^b f(t)\Delta g_2(t),
\end{eqnarray*}\]
whereupon we get \eqref{34}. This completes the proof.
Let \(g_1\) and \(g_2\) be strictly increasing functions on \(I\) and \(f\in \mathcal{R}(g_1, I)\), \(f\in \mathcal{R}(g_2, I)\). Prove that for any \(\alpha, \beta\in \mathbb{R}\), \(\alpha, \beta\geq 0\), one has \(f\in \mathcal{R}(\alpha g_1+\beta g_2, I)\).
Let \(g\) be a strictly increasing function on \(I\), \(\widetilde{T}\) be a time scale and \(\phi: \widetilde{T}\to \mathbb{R}\) be a strictly increasing continuous function, and \(\T=\phi(\widetilde{\mathbb{T}})\), and \(\phi: [A, B]\to I\) is onto, where \(A, B\in \widetilde{\mathbb{T}}\), \(A\leq B\). Let also, \(f\in \mathcal{R}(g, I)\). Then \(f\circ \phi\in \mathcal{R}(g\circ \phi, [A, B]_{\widetilde{T}})\) and
\[\begin{equation}
\label{36} \int\limits_a^b f(t)\Delta g(t)= \int\limits_A^B f(\phi(s))\widetilde{\Delta} g(\phi(s)).
\end{equation}\]
Proof
Take \(\epsilon>0\) arbitrarily. Since \(f\in \mathcal{R}(g, I)\), by the \(\epsilon\)-criterion, it follows that there is a partition \(P=\{t_0, t_1, \ldots, t_n\}\) of \(I\) such that
\[\begin{equation}
\label{38} U(P, f, g)-L(P, f, g)<\epsilon.
\end{equation}\]
Since \(\phi: [A, B]_{\widetilde{\T}}\to I\) is strictly increasing and onto, there is a partition \(Q=\{s_0, s_1, \ldots, s_n\}\) of \([A, B]_{\widetilde{\T}}\) such that
\[
\phi(s_j)=t_j,\quad j\in \{0, 1, \ldots, n\}.
\]
Thus,
\[
\phi\left([s_{j-1}, s_j]_{\widetilde{T}}\right)=[t_{j-1}, t_j],\quad j\in \{1, \ldots, n\}.
\]
Therefore
\[\begin{equation}
\label{37.1}
\begin{array}{lll}
L(P, f, g)&=& L(Q, f\circ \phi, g\circ \phi),\\ \\
U(P, f, g)&=& U(Q, f\circ \phi, g\circ \phi).
\end{array}
\end{equation}\]
Hence, applying \eqref{38}, we arrive at
\[\begin{eqnarray*}
U(Q, f\circ \phi, g\circ \phi)-L(Q, f\circ \phi, g\circ \phi)&=& U(P, f, g)-L(P, f, g)\\ \\
&<& \epsilon,
\end{eqnarray*}\]
which is the \(\epsilon\)-criterion. This proves \(f\circ \phi\in \mathcal{R}(g\circ \phi, [A, B]_{\widetilde{\mathbb{T}}})\). By \eqref{37.1}, taking supremum in the first equation over \(P\in \mathcal{P}(I)\) and \(Q\in \mathcal{P}([A, B]_{\widetilde{\mathbb{T}}})\), and taking infimum in the second equation over \(P\in \mathcal{P}(I)\) and \(Q\in \mathcal{P}([A, B]_{\widetilde{\mathbb{T}}})\),we obtain \eqref{36}. This completes the proof.