Fubini and Change of Variables: Examples and Counterexamples

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

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

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

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 $x\ne0$, 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 $-1/2$ and $1/2$.

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 $\varepsilon>0$, 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 $0$.

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 $\partial_y\exp(xy)=x\exp(xy)$, [L2] gives $\int_0^1x\exp(xy)\,dy=e^x-1$, 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 $\operatorname{cont}(S(E))=|1|\operatorname{cont}(E)=\operatorname{cont}(E)$.

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 $cx_i$, 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 $A=\begin{pmatrix}2&1&0\\0&3&1\\0&0&-2\end{pmatrix}.$

The signed-permutation formula [L2] leaves only the diagonal term, so $\det A=2\cdot3\cdot(-2)=-12$.

By [L1], the content is $|-12|=12$. 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 $DP=\begin{pmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix},\qquad \det DP=r.$

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 $(1/2,5/2)\times(\pi/12,5\pi/12)$, where $\det DP$ never vanishes. Thus the compact theorem's neighborhood hypotheses hold.

Applying [L3] to the constant-one function and integrating $r$ gives $\operatorname{area}(P(K))=\int_{\pi/6}^{\pi/3}\int_1^2r\,dr\,d\theta=\frac32\cdot\frac\pi6=\frac{\pi}{4}.$

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

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 $(1/2,5/2)\times(\pi/12,5\pi/12)\times(-2,2)$, 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 $\det DS=r^2\sin\phi$. It is positive on the parameter box by [L2].

The image norm recovers $r$; the quotient of the third coordinate by $r$ recovers $\phi$ through strict cosine monotonicity; the first two normalized coordinates then recover $\theta$. The same recovery works when both angles range in $(\pi/12,5\pi/12)$ and $r\in(1/2,5/2)$, 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 $r^2\sin\phi$.

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 $(r,\phi_1,\xi)\longmapsto(s,\rho,\xi)=(r\cos\phi_1,r\sin\phi_1,\xi) \longmapsto(s,\Phi_n(\rho,\xi)),$ where $\Phi_n$ is the dimension-$n$ hyperspherical map in the induction hypothesis.

By [L1] and [L2], the first map in step 1.2 has a $2\times2$ Jacobian block of determinant $r(\cos^2\phi_1+\sin^2\phi_1)=r$ and an identity block in $\xi$. The second has a $1\times1$ identity block and the Jacobian of $\Phi_n$ 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 $r\rho^{n-1}\prod_{j=2}^{n-1}\sin^{n-j}\phi_j =r^n\sin^{n-1}\phi_1\prod_{j=2}^{n-1}\sin^{n-j}\phi_j.$ On the stated boxes, the image norm recovers $r$, then successive coordinate ratios and strict cosine monotonicity [L4] recover every $\phi_j$, 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 $\int_0^1 1\,dy=1$.

Since $\varphi'=-1$, the proposed un-absolute right side is $\int_0^1(-1)\,dx=-1$, not $1$.

With the required absolute value, [L1] gives $\int_0^1|-1|\,dx=1$. 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, $\int_{1}^{4}1\,dy=3$, while the proposed source integral is $\int_K|2x|\,dx=\int_{-2}^{-1}(-2x)\,dx+\int_1^2 2x\,dx=3+3=6.$

The mismatch $6\ne3$ is exact double counting. Thus nonvanishing derivative does not replace injectivity, and [L1]'s hypothesis is essential.