For a continuous integrand, the Riemann-Stieltjes and Lebesgue-Stieltjes integrals agree

Extend $g$ to a continuous function $\tilde{g}:\mathbb{R}\to \mathbb{R}$ by setting $\tilde{g}(x)=g(a)$ for $x<a$, $\tilde{g}(x)=g(x)$ for $x\in [a,b]$, and $\tilde{g}(x)=g(b)$ for $x>b$. Then [L5] makes $\tilde{g}$ Borel measurable. Put $h:=(B+\tilde{g}){\chi }_{(a,b]}\mathrm{.}$ This is a nonnegative measurable function. Since $0\le h\le 2B{\chi }_{(a,b]}$, [L3], [L4], [L6], and [L7] show that $h$ is ${\mu }_F$-integrable. Let $J:={\int }_a^b(B+g)dF\mathrm{.}$ By [L1], the integral $J$ exists. The constant integrand $B$ has the same Riemann-Stieltjes sum $B(F(b)-F(a))$ for every tagged partition, so [L8] gives ${\int }_a^{b}BdF=B(F(b)-F(a))\mathrm{.}$ Therefore [L9] yields $J=I+B(F(b)-F(a))\mathrm{.}$

Let $\epsilon >0$. By [L2], choose a partition $P_0=\{a=t_0<\cdots <t_m=b\}$ with $U_F(g,P_0)-L_F(g,P_0)<\epsilon$. By [L8], choose $\delta >0$ such that every tagged partition of mesh below $\delta$ has Riemann-Stieltjes sum for $B+g$ within $\epsilon$ of $J$. Let $P=\{a=s_0<\cdots <s_n=b\}$ be a refinement of $P_0$ with mesh below $\delta$. For each $i$ put $m_i:={\inf }_{[s_{i-1},s_i]}g,M_i:={\sup }_{[s_{i-1},s_i]}g,$ and define nonnegative simple functions on $\mathbb{R}$ by ${\mathrm{\ell }}_P:={\sum }_{i=1}^n(B+m_i){\chi }_{(s_{i-1},s_i]},u_P:={\sum }_{i=1}^n(B+M_i){\chi }_{(s_{i-1},s_i]}\mathrm{.}$ Then ${\mathrm{\ell }}_P\le h\le u_P$. Because each refined infimum is at least the corresponding coarse infimum and each refined supremum is at most the corresponding coarse supremum, the Stieltjes lower sum increases and the upper sum decreases under this refinement, so $U_F(g,P)-L_F(g,P)\le U_F(g,P_0)-L_F(g,P_0)<\epsilon \mathrm{.}$ Also [L3] and [L4] give $\int {\mathrm{\ell }}_{P}d{\mu }_F=B(F(b)-F(a))+L_F(g,P),\int u_{P}d{\mu }_F=B(F(b)-F(a))+U_F(g,P)\mathrm{.}$

Fix any tagging $\xi$ of $P$. Then $\int {\mathrm{\ell }}_{P}d{\mu }_F\le \int hd{\mu }_F\le \int u_{P}d{\mu }_F,$ and because $m_i\le g({\xi }_i)\le M_i$ on every subinterval, $\int {\mathrm{\ell }}_{P}d{\mu }_F\le S(B+g,F;P,\xi )\le \int u_{P}d{\mu }_F\mathrm{.}$ Hence both $\int hd{\mu }_F$ and $S(B+g,F;P,\xi )$ lie in an interval of length $U_F(g,P)-L_F(g,P)<\epsilon$, so $\mid \int hd{\mu }_F-S(B+g,F;P,\xi )\mid <\epsilon \mathrm{.}$

Because $\mathrm{\parallel }P\mathrm{\parallel }<\delta$, [L8] gives $\mid S(B+g,F;P,\xi )-J\mid <\epsilon \mathrm{.}$ Combining this with step 2.1, $\mid \int hd{\mu }_F-J\mid <2\epsilon \mathrm{.}$ Since $\epsilon >0$ was arbitrary, $\int hd{\mu }_F=J=I+B(F(b)-F(a))\mathrm{.}$

Step 1.1 gives $h=\tilde{g}{\chi }_{(a,b]}+B{\chi }_{(a,b]}$, and both summands are integrable by step 1.1. Therefore [L3] and [L6] yield ${\int }_{(a,b]}gd{\mu }_F=\int hd{\mu }_F-B{\mu }_F((a,b])=(I+B(F(b)-F(a)))-B(F(b)-F(a))=I\mathrm{.}$ This is exactly ${\int }_{(a,b]}gd{\mu }_F={\int }_a^{b}gdF,$ so the two integrals agree.