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The indicator of the rationals in $[0, 1]$ is Lebesgue integrable with integral $0$ and not Riemann integrable

Example

Assume the Axiom of Countable Choice. Let $f : = 1_{Q \cap [0, 1]} : [0, 1] \to R$ . Then $f$ is Lebesgue integrable with $\int_{[0, 1]} f d λ_{1} = 0,$ but $f$ is not Riemann integrable on $[0, 1]$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and the Dirichlet function $1_{Q}$ .

[L1]

The Dirichlet function is the indicator of the rationals. (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ )

[L2]

Every at most countable subset of $R$ has measure zero. (Every at most countable subset of $R$ has measure zero)

[L3]

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

[L4]

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)

[L5]

The Dirichlet function is continuous at no point of $R$ . (The Dirichlet function is continuous at no point of $R$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t (c)$ )

[L6]

A bounded function on $[0, 1]$ is Riemann integrable exactly when its discontinuity set has Lebesgue measure $0$ . (A bounded function on a closed bounded interval, or on a closed nondegenerate rectangle, is Riemann integrable exactly when its discontinuity set has Lebesgue measure zero)

[L7]

The interval $[0, 1]$ has Lebesgue measure $1$ . (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)

Verification

technique · direct

1.1

By [L1], the function $f$ is $1$ on $Q \cap [0, 1]$ and $0$ on its [L1, L2, L3, L4] complement. The set $Q \cap [0, 1]$ is countable, hence null by [L2], and [L3] makes it Lebesgue measurable. Therefore $f = 0$ almost everywhere on $[0, 1]$ , so [L4] gives $\int_{[0, 1]} f d λ_{1} = 0.$

1.2

By [L5], every point of $[0, 1]$ is a discontinuity of $f$ . Thus the [L5, L6, L7] discontinuity set of $f$ is the whole interval $[0, 1]$ , whose Lebesgue measure is $1$ by [L7], not $0$ . So [L6] shows that $f$ is not Riemann integrable. ∎

Depends on

Used by

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Version 5.0, Example 9.2 (standard reference, not scraped)

The indicator of the rationals in [0,1][0,1] is Lebesgue integrable with integral 00 and not Riemann integrable