Theorem 5.4.4 (Fubini’s Theorem for Riemann-Stieltjes Integrals).label Let $[a,b],[c,d]\suf\R$, $E,F,G,H$ be a locally convex space over $K\in \curl{\R,\com}$ with $H$ sequentially complete, $E\ti F\ti G\to H$ with $(x,y,z)\mto xyz$ is a 3-linear map, $\al\in BV([a,b];F),\be\in BV([c,d];G)$ and $f\in C([a,b]\ti [c,d];E)$ then
\begin{align*}\integral{a }{b }\integral{c }{d }f(s,t)\be(dt)\al(ds)=\integral{c }{d }\integral{a }{b }f(s,t)\al(ds)\be(dt)\end{align*}
Proof. Define the continuous bilinear maps
\begin{align*}\lam:E\ti F\to L(G;H)&&\lam(x,y)\curl{z}\define xyz\\ \mu:E\ti G\to L(F;H)&&\mu(x,z)\curl{y}\define xyz\end{align*}
By Proposition 5.3.7 $f\in C([a,b]\ti [c,d];E)=UC([a,b]\ti[c,d];E)$ hence $\curl{f(\cd,t):t\in[c,d]}\suf C([a,b];E)$ is uniformly equicontinuous so by Proposition 5.4.3 we can define
\begin{align*}g:[a,b]\to L(F;H)&&s\mto \integral{c }{d }\mu\parens{f(s,t),\be(dt)}\end{align*}
then observe for any $(P=\curl{x_j}_{0}^{n},c=\curl{c_j}_{1}^{n})\in \scr{P}_{t}([a,b])$
\begin{align*}S(P,c,g,\al)&=\sums{n }{j=1}g(c_{j})\curl{\al(x_j)-\al(x_{j-1})}\\&=\sums{n }{j=1}\integral{c }{d }\mu\parens{f(s,t),\be(dt)}\curl{\al(x_j)-\al(x_{j-1})}\\&=\integral{c }{d }\sums{n }{j=1}f(c_{j},t)\parens{\al(x_j)-\al(x_{j-1})}\be(dt)\\&=\integral{c }{d }S(P_{n},c_{n},\lam(f(\cd,t),\cd)\al)\be(dt)\end{align*}
Since $\al\in BV([a,b];F)$ by Proposition 5.4.3, for any $\curl{(P_n,c_n)_{n\geq 1}}\suf \scr{P}_{t}([a,b])$ with $\s(P_{n})\to 0$
\begin{align*}\integral{a }{b }\integral{c }{d }f(s,t)\be(dt)\al(ds)=\integral{a }{b }g(s)\al(ds)=\limit{n\to\infty}S(P_{n},c_{n},g,\al)\end{align*}
By Proposition 5.3.7 $f\in C([a,b]\ti [c,d];E)=UC([a,b]\ti[c,d];E)$ hence $\curl{f(\cd,t):t\in[c,d]}\suf C([a,b];E)$ is uniformly equicontinuous so by Proposition 5.4.3
\begin{align*}\limit{n\to\infty}S(P_{n},c_{n},\lam(f(\cd,t),\cd)\al)=\integral{a }{b}\lam\parens{f(s,t),\al(ds)}\end{align*}
uniformly for all $t\in [c,d]$. As $\be\in BV([c,d];G)$ we may interchange limit and integral by Proposition 5.4.2 to obtain
\begin{align*}\limit{n\to\infty}\integral{c }{d }S(P_{n},c_{n},\lam \parens{f(\cd,t),\cd},\al)\be(dt)=\integral{c }{d }\integral{a }{b }\lam \parens{f(s,t),\al(ds)}\curl{\be(dt)}=\integral{c }{d }\integral{a }{b}f(s,t)\al(ds)\be(dt)\end{align*}
so we are done by the observation.$\square$