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 $c{\sum }_i\mathrm{vol}(Q_i)=c$.

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 $\epsilon 2^{-j-1}$; countable-union closure gives nullity.

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

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 $1\mathrm{/}2$; 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 $S\times [0,1]\subseteq \mathrm{\partial }U$.

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 $N\ge 1$ equal subintervals. On each one, $\mathrm{\mid }{x}^2-y^2\mathrm{\mid }\le 2\mathrm{/}\iota (N)$, so four squares of side $1\mathrm{/}\iota (N)$ cover that graph piece. The resulting finite cover has total area at most $4\mathrm{/}\iota (N)$, 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 $N(N-1)\mathrm{/}2$ cells and the upper staircase has $(N^2+3N-2)\mathrm{/}2$ cells.

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

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

Given $\epsilon >0$, cover $C$ by finitely many positive-width intervals $I_r$ with ${\sum }_r{\mathrm{\ell }}_r$ and ${\max }_r{\mathrm{\ell }}_r$ sufficiently small. Degenerate members may be enlarged within the budget.

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

Summing gives at most ${\sum }_r{\mathrm{\ell }}_r+2({\max }_r{\mathrm{\ell }}_r){\sum }_r{\mathrm{\ell }}_r<\epsilon$. 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.