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Change of variable for the Riemann–Stieltjes integral

Statement

Let $c < d$ and $a < b$ , and let $ϕ : [c, d] \to [a, b]$ be a strictly increasing continuous bijection. For functions $f, α : [a, b] \to R$ , one of the two Riemann–Stieltjes integrals below exists if and only if the other does, and in that case

$\int_{a}^{b} f d α = \int_{c}^{d} (f \circ ϕ) d (α \circ ϕ) .$

The nondegeneracy hypotheses are not cosmetic. If $c = d$ and $a = b$ both integrals are $0$ by the singleton convention and the identity holds trivially, but no partition exists and the argument below does not apply. If the written endpoints are reversed the intervals are empty, the empty map is vacuously such a bijection, and $f, α$ typed on an empty interval give the displayed integrals no values; that case is excluded rather than asserted.

Facts & Assumptions

Given: A strictly increasing continuous bijection $ϕ : [c, d] \to [a, b]$ and functions $f, α$ on $[a, b]$ .

[L1]

The inverse $ϕ^{- 1}$ is continuous (Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ ).

[L2]

Both $ϕ$ and $ϕ^{- 1}$ are uniformly continuous on their compact domains (Heine-Cantor in $R$ : a continuous real function on a compact subset of $R$ is uniformly continuous, proved $R$ -natively from sequential compactness).

[L3]

The Stieltjes integral is the common mesh limit of its tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L4]

A Stieltjes integral, when it exists, is unique (The Riemann–Stieltjes integral is unique).

Proof

technique · direct

1.1

If $Q = (s_{j})$ is a partition of $[c, d]$ with tags $η_{j}$ , then $P = (ϕ (s_{j}))$ is a partition of $[a, b]$ with tags $ϕ (η_{j})$ . Direct substitution gives [given] $S_{α \circ ϕ} (f \circ ϕ; Q, η) = S_{α} (f; P, ϕ \circ η) .$

2.1

By uniform continuity of $ϕ$ in [L2], arbitrarily fine $Q$ give arbitrarily fine image partitions $P$ . Thus existence of the left-hand integral in the displayed formula forces the right-hand sums to converge to the same value. Applying the identical argument to $ϕ^{- 1}$ , using [L1] and [L2], proves the converse. Uniqueness [L4] identifies the two limits.

step 1.1L1L2L3L4∎

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Sources

W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.19 (standard reference, not scraped)