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The identity integrator recovers the Riemann integral

Statement

Let $a\le b$ . For the identity function $\operatorname{id}(x)=x$ , a bounded $f:[a,b]\to\mathbb R$ is Riemann–Stieltjes integrable with respect to $\operatorname{id}$ exactly when it is Riemann integrable, and then

$\int_a^b f\,d\operatorname{id}=\int_a^b f(x)\,dx.$

For $a=b$ both sides are $0$ by the singleton conventions. For reversed endpoints the statement is about a function defined on the sorted interval: if $a>b$ and $f$ is bounded on $[b,a]$ , then both oriented conventions negate the corresponding sorted integral, so the equality is inherited from the case just proved. The hypothesis $a\le b$ is needed for the displayed clause itself, because $[a,b]$ is empty when $a>b$ and a function typed on it supplies no values to integrate.

Facts & Assumptions

Given: A bounded function $f$ on the compact interval with endpoints $a,b$ .

[L1]

A Stieltjes sum is $\sum_i f(\xi_i)(\alpha(t_{i+1})-\alpha(t_i))$ (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

A Riemann sum is $\sum_i f(\xi_i)(t_{i+1}-t_i)$ (Tagged partitions of $[a,b]$ , with a tag $\xi_i$ in each subinterval, and the Riemann sum $S(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i$ ).

[L3]

Darboux and tagged-sum Riemann integrability agree (The Darboux and Riemann definitions agree: a bounded $f$ on $[a,b]$ is Darboux integrable with integral $I$ if and only if for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that $|S(f,P,\xi) - I| < \varepsilon$ for every tagged partition of mesh below $\delta$ ).

[L4]

Oriented integrals reverse sign when endpoints are interchanged (The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

Proof

technique · direct

1.1

When $a<b$ and $\alpha=\operatorname{id}$ , every increment $\alpha(t_{i+1})-\alpha(t_i)$ equals $t_{i+1}-t_i$ . Thus [L1] and [L2] are termwise identical for every tagged partition.

L1L2

2.1

Consequently the two mesh limits exist simultaneously and have the same value; [L3] identifies that tagged limit with the Darboux integral. For $a=b$ both conventions give zero, and [L4] handles $a>b$ .

step 1.1L3L4∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 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, Definition 6.1 (standard reference, not scraped)