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ExampleConstruction: Literature-sourcedVerification: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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An integrable function on the unit square with one Dirichlet section and only one defined order of ordinary iteration

Example

On Q=[0,1]2Q=[0,1]^2, define f(x,y)={1,x=12 and yQ,0,otherwise.f(x,y)=\begin{cases}1,&x=\tfrac12\text{ and }y\in\mathbb Q,\\0,&\text{otherwise}.\end{cases} Then ff is Riemann integrable with integral 00. Every horizontal section is integrable, but the vertical section at x=1/2x=1/2 is the Dirichlet function. Hence the xx-then-yy ordinary iterated integral exists and equals 00, whereas the other ordinary order is not defined.

Facts & Assumptions

Given: The displayed function on the unit square.

[L2]

The published boundedness false statement computes the Dirichlet function's unequal lower and upper Darboux integrals, so boundedness alone does not give integrability (FALSE: every bounded function on [a,b][a,b] is Riemann integrable).

[L3]

A bounded function is Riemann integrable when grids make its upper-minus-lower sum arbitrarily small (Riemann's criterion on a nondegenerate rectangle in Rm\mathbb{R}^m: integrability is equivalent to arbitrarily small Darboux gaps).

[L4]

Riemann--Fubini distinguishes ordinary iterated integrals from lower/upper completion across exceptional sections (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

Verification

technique · direct
1.1

Every lower sum of ff is 00. Put grid lines immediately to either side of x=1/2x=1/2 with total intervening width below ε\varepsilon; only cells in that strip can have supremum 11, so the upper sum is below ε\varepsilon. Thus [L3] gives integrability and integral 00.

L3given
2.1

For fixed yy, the xx-section is either zero or the indicator of the singleton {1/2}\{1/2\}, and in both cases its integral is 00. The resulting outer function is zero, so that ordinary iteration exists and equals 00.

step 1.1
3.1

For fixed x1/2x\ne1/2, the yy-section is zero; at x=1/2x=1/2 it is the Dirichlet function and is nonintegrable by [L2]. Therefore the reverse ordinary iteration is undefined, exactly as [L4] permits despite multiple integrability.

L1L2L4step 2.1

Depends on

Used by

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Sources