Theorem 13.4.4: Fubini's Theorem for Riemann-Stieltjes Integrals

Theorem 13.4.4 (Fubini’s Theorem for Riemann-Stieltjes Integrals).label Let $[a, b], [c, d] \subset \real$, $E, F, G, H$ be a locally convex space over $K \in \RC$ with $H$ being sequentially complete, $E \times F \times G \to H$ with $(x, y, z) \mapsto xyz$ be a $3$-linear map^[1], $\alpha \in BV([a, b]; F)$, $\beta \in BV([c, d]; G)$, and $f \in C([a, b] \times [c, d]; E)$, then

\[\int_{a}^{b} \int_{c}^{d} f(s, t) \beta(dt) \alpha(ds) = \int_{c}^{d}\int_{a}^{b} f(s, t) \alpha(ds) \beta(dt)\]

Proof. Let

\[g: [a, b] \to L(F; H) \quad s \mapsto \int_{c}^{d} f(s, t) \beta(dt)\]

then for any $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_{t}([a, b])$,

\begin{align*}S(P, c, g, \alpha)&= \sum_{j = 1}^{n} g(c_{j}) [\alpha(x_{j}) - \alpha(x_{j-1})] \\&= \sum_{j = 1}^{n} \int_{c}^{d} f(c_{j}, t) \beta(dt) [\alpha(x_{j}) - \alpha(x_{j-1})] \\&= \int_{c}^{d} S(P, c, f(\cdot, t), \alpha) \beta(dt)\end{align*}

Since $f \in C([a, b] \times [c, d]; E)$, $f$ is uniformly continuous by Proposition 6.4.5, and $\bracs{f(\cdot, t)|t \in [c, d]}\subset C([a, b]; E)$ is uniformly equicontinuous. As $\alpha \in BV([a, b]; F)$, by Proposition 13.4.3, for any $\seq{(P_n, c_n)}\subset \scp_{t}([a, b])$,

\[\int_{a}^{b} \int_{c}^{d} f(s, t) \beta(dt) \alpha(ds) = \limv{n}S(P_{n}, c, g, \alpha)\]

and

\[\limv{n}S(P_{n}, c_{n}, f(\cdot, t), \alpha) = \int_{a}^{b} f(s, t) \alpha(ds)\]

uniformly for all $t \in [c, d]$. Finally, given that $\beta \in BV([c, d]; G)$,

\[\int_{c}^{d}\int_{a}^{b} f(s, t) \alpha(ds) \beta(dt) = \limv{n}\int_{c}^{d} S(P_{n}, c_{n}, f(\cdot, t), \alpha) \beta(dt)\]

by Proposition 13.4.2.$\square$

$E, F, G$ are assumed to be disjoint, so the product is well-defined regardless of the order of the terms.keyboard_return

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