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A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral

Statement

Assume the Axiom of Countable Choice. Let $a < b$ and let $f : [a, b] \to R$ be bounded and Riemann integrable. Then $f$ is Lebesgue measurable on $[a, b]$ and is integrable there, and its Lebesgue integral equals its Riemann integral: $\int_{[a, b]} f d λ_{1} = \int_{a}^{b} f (x) d x .$

This is the point at which the completeness of Lebesgue measure is used essentially: the proof obtains a Borel function equal to $f$ almost everywhere, and measurability of $f$ itself is then a completeness statement.

Facts & Assumptions

Given: The Axiom of Countable Choice, reals $a < b$ , a bounded Riemann integrable function $f : [a, b] \to R$ , its Riemann integral $I : = \int_{a}^{b} f (x) d x$ , and a real $B > 0$ with $∣ f (x) ∣ \leq B$ on $[a, b]$ .

[L1]

The envelope lemma produces bounded Borel functions $φ, ψ : [a, b] \to R$ with $φ \leq f \leq ψ$ and $\int_{[a, b]} φ d λ_{1} = I = \int_{[a, b]} ψ d λ_{1} .$ (A bounded Riemann integrable function admits Borel Darboux envelopes with the same Lebesgue integral)

[L2]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

[L3]

On a complete measure space, a function equal almost everywhere to a measurable function is measurable. (On a complete measure space, equality almost everywhere preserves measurability)

[L4]

Lebesgue measure on $R$ is complete. (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume)

[L5]

Two integrable functions that agree almost everywhere have the same integral over every measurable set. (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree)

[L6]

If a real measurable function is bounded in absolute value by a nonnegative integrable function, then its absolute value has finite integral; a measurable real function is integrable exactly when the integral of its absolute value is finite. (Closure properties of measurable functions used by the integral, Monotonicity and nonnegative homogeneity of the nonnegative integral, The nonnegative integral agrees with the simple integral on simple functions, The integral of a nonnegative simple function, Integrable real and complex functions, and their integrals)

[L7]

The interval $[a, b]$ is Lebesgue measurable with measure $b - a$ . (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included)

Proof

technique · direct

1.1L1L2

By [L1], choose bounded Borel functions $φ, ψ$ on $[a, b]$ with [L1, L2] $φ \leq f \leq ψ$ and both integrals equal to $I$ . Then $0 \leq ψ - φ$ and $\int_{[a, b]} (ψ - φ) d λ_{1} = 0.$ So [L2] gives $ψ = φ$ almost everywhere. Since $φ \leq f \leq ψ$ , the same null set yields $f = φ$ almost everywhere.

2.1L4step 1.1L3

By [L4], the measure space $(R, L (R), λ_{1})$ is [step 1.1, L3, L4] complete. The function $φ$ is measurable because it is Borel, so [L3] applied to step 1.1 shows that $f$ is Lebesgue measurable.

3.1step 2.1L1L6L7

The constant function $B χ_{[a, b]}$ is a nonnegative simple measurable [step 2.1, L1, L6, L7] function, and [L6] together with [L7] gives $\int_{[a, b]} B χ_{[a, b]} d λ_{1} = B (b - a) < + \infty .$ Since step 2.1 makes $∣ f ∣$ measurable and $∣ f ∣ \leq B χ_{[a, b]}$ , [L6] yields $\int_{[a, b]} ∣ f ∣ d λ_{1} < + \infty .$ Hence $f$ is Lebesgue integrable on $[a, b]$ . The same estimate applies to $φ$ , because [L1] gives $∣ φ ∣ \leq B$ .

4.1step 1.1step 3.1L1L5∎

Steps 1.1 and 3.1 show that $f$ and $φ$ are integrable and agree [step 1.1, step 3.1, L1, L5] almost everywhere. Taking the measurable set $A = [a, b]$ in [L5] gives $\int_{[a, b]} f d λ_{1} = \int_{[a, b]} φ d λ_{1} .$ By [L1], the right-hand side is $I = \int_{a}^{b} f (x) d x$ . So the Lebesgue and Riemann integrals of $f$ agree.

Depends on

Used by

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Characteristic function of a gaussian law Example
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Independent sums via characteristic functions Example
Ito formula for Brownian powers Example
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Legendre polynomials from Gram–Schmidt Example
Matching C¹ pieces across a hyperplane have no jump derivative Example
Odd reflection at a Dirichlet endpoint Example
One-dimensional Dirichlet Green kernel on an interval Example
Poincare on an interval: the length dependence is linear Example
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Strong law estimator of an integrable mean Example
The absolute value has a weak first derivative Example
The Fourier series of a sawtooth and the Basel sum Example

…and 36 more results.

Dependency tree · two levels

54 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, Theorem 9.1 (standard reference, not scraped)
Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis, Theorem (5.52) (standard reference, not scraped)