Is There a Compact Set That Is Not Jordan Measurable?

Yes. Compactness does not guarantee Jordan measurability. A useful example is the Smith–Volterra–Cantor set, also called the fat Cantor set: it is a compact subset of the real line with positive measure but empty interior.

The key point is that a bounded set is Jordan measurable exactly when its boundary has Lebesgue measure zero. For this example, the boundary is the entire set, and that set has measure $1/2$.

The boundary criterion for Jordan measurability

A bounded set $A\subseteq\mathbb{R}^n$ is Jordan measurable if its indicator function is Riemann integrable. The Lebesgue criterion for Riemann integrability says this happens precisely when the set of discontinuities of the indicator function has measure zero. Those discontinuities are exactly the boundary points of $A$. Thus, for bounded sets,

$$A\text{ is Jordan measurable}\quad\Longleftrightarrow\quad m(\partial A)=0,$$

where $m$ denotes Lebesgue measure and $\partial A$ is the boundary of $A$.

This criterion shows why compactness alone is not enough: compact sets are closed and bounded, but their boundaries can have positive measure.

Constructing a compact example

Begin with $K_0=[0,1]$. At stage $n$, remove an open interval of length $1/4^n$ from the middle of each interval remaining from stage $n-1$. Let $K_n$ be the set left after stage $n$, and define

$$K=\bigcap_{n=0}^{\infty}K_n.$$

At stage $n$, there are $2^{n-1}$ intervals to remove. The total length removed over all stages is therefore

$$\begin{aligned} \sum_{n=1}^{\infty}\frac{2^{n-1}}{4^n} &=\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}\\ &=\frac{1}{2}. \end{aligned}$$

The removed intervals are disjoint, so the remaining set has Lebesgue measure

$m(K)=1-\frac12=\frac12.$

The construction is possible at every stage: each interval being split is longer than the middle interval removed from it. For instance, at the first stage, removing an interval of length $1/4$ leaves two intervals of length $3/8$.

Why the set is compact but not Jordan measurable

Each $K_n$ is a finite union of closed intervals, so it is closed and bounded. The intersection $K$ is also closed and bounded. By the Heine–Borel theorem, $K$ is compact.

The lengths of the remaining intervals tend to zero as the construction continues. Therefore $K$ cannot contain an interval of positive length: any such interval would have to fit inside one of the remaining intervals at every stage, even though those intervals become arbitrarily short. Hence $K$ has empty interior. Since it is closed,

$\partial K=K.$

It follows that $m(\partial K)=m(K)=1/2$, which is not zero. By the boundary criterion, $K$ is not Jordan measurable. This gives a compact set that is not Jordan measurable, even though it is Lebesgue measurable and has finite measure.

Common mistakes

• Assuming closed and bounded implies Jordan measurable. In $\mathbb{R}^n$, closed and bounded implies compact, not that the boundary has measure zero.
• Confusing Jordan measurability with Lebesgue measurability. The fat Cantor set is Lebesgue measurable. Jordan measurability imposes the additional boundary-zero condition.
• Using the ordinary Cantor set as the example. The usual middle-thirds Cantor set is compact, but it has measure zero and is Jordan measurable. The fat Cantor construction is chosen to leave positive measure while still having empty interior.
• Thinking empty interior means measure zero. A set can have no interval inside it and still have positive measure, as this example demonstrates.