The Riemann Integral in R^m and Jordan Content: Examples and Counterexamples

Every side has length $1$, so the finite product volume is $1$.

On every cell, both infimum and supremum of the constant function are $c$. Thus both sums are $c\sum_i\operatorname{vol}(Q_i)=c$.

Lower and upper integrals therefore both equal $c$.

The set $E$ is countable by [L1]. Enumerate its points and cover the $j$-th singleton by a square with volume below $\varepsilon2^{-j-1}$; countable-union closure gives nullity.

By [L2], both $E$ and its complement are dense at every point of the unit square, so $\partial E=[0,1]^2$.

The unit square has content $1$, by The unit box in $\mathbb{R}^m$ has volume $1$, and the integral of a constant $c$ over it is $c$ and A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content, hence is not null (For compact subsets of $\mathbb{R}^m$, measure zero and content zero coincide).

The boundary criterion [L3] makes $E$ non-Jordan-measurable, despite nullity.

The slab $K$ is closed and bounded, hence compact by Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. Since $S$ has empty interior, $K$ has empty interior and, being closed, equals its boundary.

By [L1] and [L2], every rectangle cover of $K$ has total area at least $1/2$; its boundary therefore does not have content zero.

The boundary criterion A bounded set in $\mathbb{R}^m$ is Jordan measurable iff its boundary is null, equivalently of content zero makes $K$ non-Jordan-measurable.

The set $(0,1)\setminus S$ is open, so $U$ is open and bounded.

Every neighbourhood of a point of $S\times[0,1]$ meets $U$, by nowhere density in the first coordinate and the interval factor, and meets the complement. Hence $S\times[0,1]\subseteq\partial U$.

Outer content is monotone under inclusion, so [L2] gives the boundary positive outer content.

By A bounded set in $\mathbb{R}^m$ is Jordan measurable iff its boundary is null, equivalently of content zero, $U$ is not Jordan measurable.

Apply [L2] to the continuous polynomial in [L1].

Directly, divide $[0,1]$ into $N\ge1$ equal subintervals. On each one, $|x^2-y^2|\le2/\iota(N)$, so four squares of side $1/\iota(N)$ cover that graph piece. The resulting finite cover has total area at most $4/\iota(N)$, which can be made arbitrarily small.

Both arguments establish content zero.

Index the grid cells by $0\le i,j<N$. A cell is contained in $T$ when $i+j\le N-2$, while it meets $T$ when $i+j\le N$. Thus the lower staircase has $N(N-1)/2$ cells and the upper staircase has $(N^2+3N-2)/2$ cells.

Induction gives $\sum_{k<N}\iota(k)=\iota(N)\iota(N-1)/2$. Hence the lower area is $(N-1)/(2N)$, the upper area is $1/2+3/(2N)-1/N^2$, and their gap is $2/N-1/N^2<2/N$. Both areas tend to $1/2$.

Steps 1.1 and 1.2 give inscribed and covering grid approximations converging to $1/2$, so the inner and outer contents agree at $1/2$. Alternatively, [L2] gives Jordan measurability from the boundary criterion. In either route, [L1] identifies the common value with the indicator integral.

Given $\varepsilon>0$, cover $C$ by finitely many positive-width intervals $I_r$ with $\sum_r\ell_r$ and $\max_r\ell_r$ sufficiently small. Degenerate members may be enlarged within the budget.

Above $I_r$, stack squares of side $\ell_r$. By [L2], at most $1/\ell_r+2$ squares suffice, with total area at most $\ell_r+2\ell_r^2$.

Summing gives at most $\sum_r\ell_r+2(\max_r\ell_r)\sum_r\ell_r<\varepsilon$. Thus the slab has cube-content zero.

By Jordan inner and outer content and Jordan measurable bounded sets in $\mathbb{R}^m$, cube-content zero makes the Jordan outer content $0$. The nonnegative inner content is at most the outer content, so both are $0$; the slab is Jordan measurable with content $0$, unlike the fat-Cantor slab The Smith–Volterra–Cantor slab $S\times[0,1]$ is compact and not Jordan measurable.