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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 aR and let f:[a,)[0,) be Riemann integrable on every compact interval [a,R] with R>a. If the improper Riemann integral af(x)dx converges in the sense of Improper integrals over unbounded intervals, then f is Lebesgue integrable on [a,) and [a,)fdλ1=af(x)dx.

Facts & Assumptions

Given: The Axiom of Countable Choice, a real a, a nonnegative function f:[a,)[0,) that is Riemann integrable on every [a,R] with R>a, and a finite improper Riemann integral L:=af(x)dx.

[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,) is the limit of the truncated Riemann integrals as the right endpoint tends to +. (Improper integrals over unbounded intervals)

Proof

technique · direct
1.1

For each natural number n1, define fn:=fχ[a,a+n]. Then 0fnfn+1 and fn(x)f(x) for every xa. By [L1] applied on [a,a+n], [a,)fndλ1=[a,a+n]fdλ1=aa+nf(x)dx.

L1construct
2.1

Since fnf, [L2] gives [a,)fdλ1=limn[a,)fndλ1=limnaa+nf(x)dx. Because a+n+, [L3] identifies the last limit with the given improper integral L. Hence [a,)fdλ1=L, and in particular the Lebesgue integral is finite, so f is integrable on the half-line.

step 1.1L2L3

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