TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28

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A nonnegative improper Riemann integral on a half-line agrees with the Lebesgue integral

Statement

Assume the Axiom of Countable Choice. Let $a \in R$ and let $f : [a, \infty) \to [0, \infty)$ be Riemann integrable on every compact interval $[a, R]$ with $R > a$ . If the improper Riemann integral $\int_{a}^{\infty} f (x) d x$ converges in the sense of Improper integrals over unbounded intervals, then $f$ is Lebesgue integrable on $[a, \infty)$ and $\int_{[a, \infty)} f d λ_{1} = \int_{a}^{\infty} f (x) d x .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a real $a$ , a nonnegative function $f : [a, \infty) \to [0, \infty)$ that is Riemann integrable on every $[a, R]$ with $R > a$ , and a finite improper Riemann integral $L : = \int_{a}^{\infty} f (x) d x$ .

[L1]

On every compact interval, a bounded Riemann integrable function is Lebesgue integrable there with the same value. (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral)

[L2]

Monotone convergence holds for nonnegative measurable functions. (Monotone convergence for the integral)

[L3]

The improper integral over $[a, \infty)$ is the limit of the truncated Riemann integrals as the right endpoint tends to $+ \infty$ . (Improper integrals over unbounded intervals)

Proof

technique · direct

1.1L1construct

For each natural number $n \geq 1$ , define $f_{n} : = f χ_{[a, a + n]} .$ Then $0 \leq f_{n} \leq f_{n + 1}$ and $f_{n} (x) ↑ f (x)$ for every $x \geq a$ . By [L1] applied on $[a, a + n]$ , $\int_{[a, \infty)} f_{n} d λ_{1} = \int_{[a, a + n]} f d λ_{1} = \int_{a}^{a + n} f (x) d x .$

2.1step 1.1L2L3∎

Since $f_{n} ↑ f$ , [L2] gives $\int_{[a, \infty)} f d λ_{1} = \lim_{n \to \infty} \int_{[a, \infty)} f_{n} d λ_{1} = \lim_{n \to \infty} \int_{a}^{a + n} f (x) d x .$ Because $a + n \to + \infty$ , [L3] identifies the last limit with the given improper integral $L$ . Hence $\int_{[a, \infty)} f d λ_{1} = L,$ and in particular the Lebesgue integral is finite, so $f$ is integrable on the half-line.

Depends on

Used by

Smooth and integrable does not imply Schwartz Counterexample
The critical Riesz potential can diverge and be essentially unbounded Counterexample
The standard intertwining operator A(nu) Definition
Hilbert transform of the line Poisson kernel Example
Poisson kernel transform and Abel summability on the line Example
The positive-type Gaussian on the real line and its cyclic model Example
The upper half-plane Poisson boundary density Example
Euclidean Gaussian transform with the 2π normalization Lemma
The sine integral under Countable Choice: uniform bounds and the value pi/2 Lemma

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard F. Bass, Real Analysis for Graduate Students, Version 5.0, Exercise 9.4 (standard reference, not scraped)
Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis, Theorem (5.53) (standard reference, not scraped)