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The total-variation bound for a Riemann–Stieltjes integral

Statement

Suppose $\int_a^b f\,d\alpha$ exists, $\alpha$ has bounded variation, and $|f(x)|\le M$ on $[a,b]$ . Then

$\left|\int_a^b f\,d\alpha\right|\le M\operatorname{Var}_{[a,b]}(\alpha).$

Facts & Assumptions

Given: An existing Stieltjes integral, a BV integrator $\alpha$ , and a bound $|f|\le M$ .

[L1]

Stieltjes sums converge to the integral in the mesh sense (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, The Riemann–Stieltjes integral is unique).

[L2]

Every sum of absolute integrator increments is bounded by total variation (Bounded variation and total variation on an interval).

[L3]

Finite sums and the triangle inequality give $|\sum u_i|\le\sum|u_i|$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).

[L4]

Non-strict inequalities pass to limits (Limits preserve non-strict inequalities).

Proof

technique · direct

1.1

Every tagged sum satisfies $|S(f,\alpha;P,\xi)|\le\sum_{i<n}|f(\xi_i)||\Delta_i\alpha|\le M\sum_{i<n}|\Delta_i\alpha|\le M\operatorname{Var}_{[a,b]}(\alpha)$ .

L2L3

2.1

Take a sequence of tagged partitions with mesh tending to $0$ . Their sums converge to the integral by [L1], and [L4] passes the bound in step 1.1 to the limit. Orientation and the singleton case preserve the same absolute-value inequality.

step 1.1L1L4∎

Depends on

Used by

A countable pure-step integrator evaluates a continuous integrand as the absolutely convergent weighted sum of its values at the jumps Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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