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A Riemann-integrable Thomae-type function whose $x$ -sections are nonintegrable at every rational height

Example

Let $t:[0,1]\to[0,1]$ be Thomae's function and define $f(x,y)=\mathbf1_{\mathbb Q}(x)t(y)\qquad((x,y)\in[0,1]^2).$ Then $f$ is Riemann integrable with integral $0$ , although its $x$ -section is nonintegrable at every rational height $y$ . Those exceptional heights are dense and are not a content-zero set. The lower and upper $x$ -section envelopes are nevertheless $0$ and $t$ , and both have integral $0$ .

Facts & Assumptions

Given: The Thomae function $t$ and the displayed product on the unit square.

[L1]

The rationals and irrationals are dense; Thomae's function is positive at a rational in inverse proportion to its least denominator and is zero at every irrational (The Dirichlet function $1_{\mathbb{Q}}$ , and Thomae's function $t$ with $t(x) = 1/q$ at a rational $x = p/q$ in lowest terms with $q \ge 1$ and $t(x) = 0$ at every irrational $x$ ).

[L2]

Product-grid Darboux gaps characterize multidimensional Riemann integrability (Riemann's criterion on a nondegenerate rectangle in $\mathbb{R}^m$ : integrability is equivalent to arbitrarily small Darboux gaps).

[L3]

Riemann--Fubini identifies the multiple integral with both lower and upper section-envelope integrals without requiring every section to be integrable (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

Verification

technique · direct

1.1

Given $\varepsilon>0$ , only finitely many reduced rationals have Thomae height at least $\varepsilon/2$ . Isolating those points in intervals of total length below $\varepsilon/2$ proves directly that $t$ is Riemann integrable with integral $0$ ; using the same intervals as horizontal strips gives a product-grid Darboux gap below $\varepsilon$ for $f$ . Hence [L2] makes $f$ integrable with integral $0$ .

L1L2

2.1

At rational height $y$ , $t(y)>0$ and the $x$ -section is a nonzero multiple of the Dirichlet function, hence nonintegrable; at irrational height it is zero. The exceptional rational heights are dense and not content zero: any finite union of closed intervals covering them also covers their closure $[0,1]$ and has total length at least $1$ . The lower and upper section integrals are $0$ and $t(y)$ , whose outer integrals both vanish by [L1] and [L3].

L1L3step 1.1algebra

3.1

In the other direction, a rational $x$ gives the section $t$ and an irrational $x$ gives zero; step 1.1 makes every such section integrable with value $0$ . Hence that ordinary iterated integral is $0$ , whereas the $x$ -first ordinary iteration is undefined at a dense set of heights.

L1step 1.1step 2.1∎

Depends on

Used by

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Sources

J. Lebl, Basic Analysis II, Exercise 10.2.9 (standard reference, not scraped)

A Riemann-integrable Thomae-type function whose xxx-sections are nonintegrable at every rational height