# Jordan outer content and Lebesgue measure coincide on compact sets: Folland's proof.

Folland's proof goes as follows: if $K$ is compact, we may find a cube $Q_0$ such that $K\subset \text {int} Q_0\subseteq Q_0.$ Then $m(Q_0)$ is expressed as a sum of the volumes of unions of dyadic cubes of two kinds: ones that intersect $K$ and ones that are contained in $\text {int} Q_0\setminus K.$ On taking limits and using the fact that the inner Jordan content of $\text {int} Q_0\setminus K$ is the same as its Lebesgue measure (because it is open), the result follows.

I would do it another way, which seems simpler and does not use the fact that inner Jordan content and the Lebesgue measure coincide on open sets. So I guess my proof is not correct. But I do not see my error:

It is clear by definition of the Jordan content and the Lebesgue measure, that $m(K)\le \overline \kappa (K).$ We prove the reverse inequality.

Let $K\subset \Bbb R^n$ be compact and $\epsilon>0.$ Of course, $K$ is Lebesgue measurable, so there is a countable collection $\left \{ R_i \right \}^{\infty}_{i=1}$ of open boxes such that $\bigcup_n R_n\supset K$ with $\sum_im(R_i)<m(K)+\epsilon.$

$K$ is compact so we may assume that the $R_i$ are bounded. And because there is a finite subcollection $\left \{ R_i \right \}^{n}_{i=1}$ such that $\bigcup^{n}_{i=1} R_i\supset K$, we now have

$\overline \kappa (K)\le \sum ^{n}_{i=1}m(\overline R_i)=\sum ^{n}_{i=1}m(R_i)\le \sum_im(R_i)<m(K)+\epsilon,$ from which it follws that $\overline \kappa (K)\le m(K).$

It seems fine to me. There is only a bit of inaccuracy in the very last assertion that $\overline \kappa (K)\le m(K)$.

You have shown that $\forall \varepsilon > 0,\ m(K) \le \overline \kappa (K) < m(K) + \varepsilon$ and this directly implies that $\overline \kappa (K) = m(K)$. You are squeezing $\overline \kappa (K)$ against $m(K)$ by above.

So of course $\overline \kappa (K)\le m(K)$ is true since $\overline \kappa (K) = m(K)$ is always true but $\overline \kappa (K) < m(K)$ never holds in this case.

On a side note I would rewrite

$K$ is compact so we may assume that the $R_i$ are bounded. And because there is a finite subcollection $\left \{ R_i \right \}^{n}_{i=1}$ such that $\bigcup^{n}_{i=1} R_i\supset K$, we now have

as something like

Since $K$ is compact, therefore $m(K) < \infty$, it follows that each $R_i$ also has finite Lebesgue measure, hence all bounded, and that there is a finite subcollection $\left \{ R_i \right \}^{n}_{i=1}$ such that $\bigcup^{n}_{i=1} R_i\supset K$. We now have

to make clearer what is going on here.