Fubini and Change of Variables: Examples and Counterexamples

Every lower sum of $f$ is $0$. Put grid lines immediately to either side of $x=1\mathrm{/}2$ with total intervening width below $\epsilon$; only cells in that strip can have supremum $1$, so the upper sum is below $\epsilon$. Thus [L3] gives integrability and integral $0$.

For fixed $y$, the $x$-section is either zero or the indicator of the singleton $\{1\mathrm{/}2\}$, and in both cases its integral is $0$. The resulting outer function is zero, so that ordinary iteration exists and equals $0$.

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

For each fixed $y$, the $x$-section is either $x\mapsto x$ or zero; both have integral $0$ on $[-1,1]$. Hence the $x$-first iterated integral exists and is $0$.

For fixed $x\mathrm{\ne }0$, density in [L1] makes every lower and upper Darboux sum of the $y$-section equal to the interval length times $\min (x,0)$ and $\max (x,0)$ respectively. Thus the section is nonintegrable, and integrating its lower and upper integral functions in $x$ gives $-1\mathrm{/}2$ and $1\mathrm{/}2$.

If $f$ were multiply Riemann integrable, [L2] would force those two envelope integrals to agree. They do not, so $f$ is not Riemann integrable although the iteration in step 1.1 exists. This refutes the claim.

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

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].

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.

By continuity and [L1], reverse the order and integrate in $y$ first. Since ${\mathrm{\partial }}_y\exp (xy)=x\exp (xy)$, [L2] gives ${\int }_0^{1}x\exp (xy)dy=e^x-1$, including $x=0$.

A second application of [L2] gives ${\int }_{-1}^1(e^x-1)dx=(e-e^{-1})-2.$

Using [L3] for the negative endpoint yields the stated value $e-e^{-1}-2$.

The shear matrix is the identity with one off-diagonal entry $c$, so its determinant is $1$. By [L2] the linear image $S(E)$ is Jordan measurable and $\mathrm{cont}(S(E))=\mathrm{\mid }1\mathrm{\mid }\mathrm{cont}(E)=\mathrm{cont}(E)$.

Sections show the same thing wherever [L1] applies. Hold all coordinates except $x_j$ fixed: the corresponding section of $S(E)$ is the section of $E$ translated by $c{x}_i$, so its one-dimensional content is unchanged, and for a set whose sections are Jordan outside a content-zero parameter set [L1] integrates these equal values to the same total. This is a second reading of the result and not a second proof of it: a bounded Jordan set need not have Jordan sections outside a content-zero parameter set, so [L1] is not available for an arbitrary $E$ and step 1.1 carries the statement.

Put the vectors into the columns of the upper-triangular matrix $A=\left(\begin{array}{ccc}2& 1& 0\\ 0& 3& 1\\ 0& 0& -2\end{array}\right)\mathrm{.}$

The signed-permutation formula [L2] leaves only the diagonal term, so $\det A=2\cdot 3\cdot (-2)=-12$.

By [L1], the content is $\mathrm{\mid }-12\mathrm{\mid }=12$. Directly, the first two vectors span a base of area $6$ in the horizontal plane and the third has perpendicular height $2$, again giving $12$.

Differentiation and [L1] give $DP=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right),\det DP=r\mathrm{.}$

Equality of two images first gives equality of radii by [L1], then equality of cosines; [L2] gives equality of angles. The same recovery works on the open neighborhood $(1\mathrm{/}2,5\mathrm{/}2)\times (\pi \mathrm{/}12,5\pi \mathrm{/}12)$, where $\det DP$ never vanishes. Thus the compact theorem's neighborhood hypotheses hold.

Applying [L3] to the constant-one function and integrating $r$ gives $\mathrm{area}(P(K))={\int }_{\pi \mathrm{/}6}^{\pi \mathrm{/}3}{\int }_1^{2}rdrd\theta =\frac{3}{2}\cdot \frac{\pi }{6}=\frac{\pi }{4}\mathrm{.}$

The derivative matrix is block triangular over the polar block, and [L1] gives $\det DC=\det \left(\begin{array}{ccc}\cos \theta & -r\sin \theta & 0\\ \sin \theta & r\cos \theta & 0\\ 0& 0& 1\end{array}\right)=r\mathrm{.}$

The third image coordinate recovers $z$. The first two recover $r$ from their Euclidean norm and then $\theta$ from strict cosine monotonicity [L2]. The same recovery works on $(1\mathrm{/}2,5\mathrm{/}2)\times (\pi \mathrm{/}12,5\pi \mathrm{/}12)\times (-2,2)$, where $r>0$, so the map is injective with invertible derivative on an open neighborhood of the compact box.

Therefore [L3] applies on this seam-free compact box, and every transformed volume integral carries precisely the factor $r$.

Differentiating the three coordinates and expanding by columns, then using [L1], gives $\det DS=r^2\sin \varphi$. It is positive on the parameter box by [L2].

The image norm recovers $r$; the quotient of the third coordinate by $r$ recovers $\varphi$ through strict cosine monotonicity; the first two normalized coordinates then recover $\theta$. The same recovery works when both angles range in $(\pi \mathrm{/}12,5\pi \mathrm{/}12)$ and $r\in (1\mathrm{/}2,5\mathrm{/}2)$, so $S$ is injective with nonzero determinant on an open neighborhood of the compact box.

The positive determinant and step 2.1 verify every hypothesis of [L3], so spherical integration on this box carries the factor $r^2\sin \varphi$.

For $n=2$, the formula is the polar determinant $r$, with the empty product of sine factors equal to $1$. Direct differentiation verifies it, while the image norm and strict cosine monotonicity recover the radius and seam-free angle.

Assume the formula in dimension $n$. Factor the dimension-$(n+1)$ coordinate map as $(r,{\varphi }_1,\xi )⟼(s,\rho ,\xi )=(r\cos {\varphi }_1,r\sin {\varphi }_1,\xi )⟼(s,{\mathrm{\Phi }}_n(\rho ,\xi )),$ where ${\mathrm{\Phi }}_n$ is the dimension-$n$ hyperspherical map in the induction hypothesis.

By [L1] and [L2], the first map in step 1.2 has a $2\times 2$ Jacobian block of determinant $r({\cos }^2{\varphi }_1+{\sin }^2{\varphi }_1)=r$ and an identity block in $\xi$. The second has a $1\times 1$ identity block and the Jacobian of ${\mathrm{\Phi }}_n$ at radius $\rho$; [L3] will discharge the induction after this step. The signed-permutation formula gives these block determinants, and multiplicativity with the induction hypothesis gives $r{\rho }^{n-1}{\prod }_{j=2}^{n-1}{\sin }^{n-j}{\varphi }_j=r^n{\sin }^{n-1}{\varphi }_1{\prod }_{j=2}^{n-1}{\sin }^{n-j}{\varphi }_j\mathrm{.}$ On the stated boxes, the image norm recovers $r$, then successive coordinate ratios and strict cosine monotonicity [L4] recover every ${\varphi }_j$, and the final planar pair recovers $\theta$. The sine factors do not vanish, so the map is injective with nonzero determinant. This proves the formula and box claim in dimension $n+1$, and [L3] completes the induction.

For every $r>0$, periodicity [L1] gives $P(r,0)=P(r,2\pi )$ although the two parameter points differ. Thus the two closed seam faces already destroy injectivity.

At $r=0$, every angle maps to the origin, providing infinitely many preimages even away from comparing the seam endpoints.

Direct differentiation and [L2] give $\det DP(r,\theta )=r$. Thus $\det DP(0,\theta )=0$, so the derivative is singular along the entire zero-radius edge and the map violates both claimed hypotheses.

The image is $[0,1]$, so its unoriented length integral is ${\int }_0^{1}1dy=1$.

Since ${\phi }^{\mathrm{\prime }}=-1$, the proposed un-absolute right side is ${\int }_0^1(-1)dx=-1$, not $1$.

With the required absolute value, [L1] gives ${\int }_0^1\mathrm{\mid }-1\mathrm{\mid }dx=1$. The negative value in step 2.1 instead belongs to the oriented formula [L2], whose image endpoints occur in reverse order. Hence the claim is false.

The derivative $2x$ never vanishes anywhere on $U$, but $g(-x)=g(x)$, so every point of $(1,4)$ has one preimage in each component of $K$.

The image integral is its length, ${\int }_1^{4}1dy=3$, while the proposed source integral is ${\int }_K\mathrm{\mid }2x\mathrm{\mid }dx={\int }_{-2}^{-1}(-2x)dx+{\int }_1^{2}2xdx=3+3=6.$

The mismatch $6\mathrm{\ne }3$ is exact double counting. Thus nonvanishing derivative does not replace injectivity, and [L1]'s hypothesis is essential.