Let $M_k$ be an elementary set, $|M_k|$ - it's volume. $K$ is a cube of volume 1. Let $E \subset K$.
Then inner Jordan measure $\mu^J_*(E)$ of set $E$ is defined as
- $\mu^J_*(E) = 1 - \mu^*_J(\overline{\rm E}) = 1 - \inf_{\cup_{k=1}^r M_k\supset \overline{\rm E}} \sum_{k=1}^r|M_k|$
- $\mu^J_*(E) = \sup_{\cup_{k=1}^r M_k\subset E} \sum_{k=1}^r|M_k|$
$E$ is Jordan-mesaurebale if $\mu^J_*(E) = \mu^*_J(E)$
Why are they equivalent? (It seems to be true, but I can't prove). And why doesn't it work for Lebesgue?
Take a collection of $M_k$s inside $E$. Fix $\varepsilon>0$. For each $M_k=[a_k^1, b_k^1]\times ... \times [a_k^d, b_k^d]$ let $N_k=(a_k^1+\varepsilon, b_k^1-\varepsilon)\times ... \times (a_k^d+\varepsilon, b_k^d-\varepsilon)$. The $N_k$s are also inside $E$. Then the complement $K\setminus \cup N_k$ covers $\bar{E}=K\setminus E$, and is an elementary set (which fails if we use countable collections instead of finite ones, so does not work for Lebesgue measure), and has volume $1-\sum|M_k|+\delta(\varepsilon)$ i.e. arbitrary close to $1-\sum|M_k|$. This means that 2 is less than or equal to 1.
The other direction is similar (start with $M_k$s covering $E$'s complement, take open bricks of slightly bigger size; their complement is elementary, is contained in $E$ and has volume essentially 1 minus volume of $M_k$s. So 1 is less than or equal to 2).
For counterexample in Lebesgue case, take $E$ to be irrationals in $[0,1]$. Then (2) is 0, and (1) is 1.