This question has a number of parts. Let $E\subset\mathbb{R}^{d}$ be a bounded subset.

(1) Show that $m^{\star,(J)}(E)=m^{\star,(J)}(\bar{E})$ (closure)

(2) Show that $m_{\star,(J)}(E)=m_{\star,(J)}(E^{\circ})$ (interior)

(3) Show that $E$ is Jordan measurable if and only if $m^{\star}(\partial E)=0$.

I originally had different proofs for parts (1) and (2), but I revised them because I wanted a more rigorous $\epsilon$-estimate type proof (I'm trying to practice getting better at this). But, looking at them again, I think there is a flaw. In particular, where I make the justification that we can "fatten up" or "shrink down" the covers in each part (see the proofs). Basically, the reason why I think it is now flawed is because if one replaces $\frac{\epsilon}{N}$ by $\frac{\epsilon}{2^{n}}$, the proof for the outer Jordan measure (part (1)) seems to carry over word-for-word in the case of Lebesgue outer measure, and that's clearly false since $\mu^{\star}([0,1]\cap\mathbb{Q})=0$, yet $\mu^{\star}(\overline{[0,1]\cap\mathbb{Q}})=1$.

Anyway, I would appreciate anyone's assistance in helping me correct these proofs. Also, for part (3), I'm stuck on proving the implication $m^{\star,(J)}(\partial E)=0\rightarrow E\in\mathscr{J}(\mathbb{R}^{d})$.

SOLUTION (1) By definition of $m^{\star,(J)}$, there exists a collection of boxes $\{B_{j}\}_{j=1}^{N}$ such that both \begin{align*} E\subset\bigcup\limits_{j=1}^{N}B_{j} &&\text{and} &&\sum\limits_{j=1}^{N}|B_{j}|\leq m^{\star,(J)}(E)+\epsilon \end{align*} hold for every $\epsilon>0$. Because $E\Delta\bar{E}\subset E'$ (the set of limit points of $E$), by enlarging each box $B_{j}$ (as necessary), we can find a new collection of boxes $\{B'_{j}\}_{j=1}^{N}$ such that each of the conditions \begin{align*} |B_{j}|\leq|B'_{j}|\leq|B_{j}|+\frac{\epsilon}{N}, &&\bigcup_{j=1}^{N}B'_{j}\supset\bar{E}\supset E, &&\text{and} &&\sum\limits_{j=1}^{N}|B'_{j}|\leq m^{\star,(J)}(\bar{E})+\epsilon \end{align*} also holds for the same $\epsilon$. Montonicity and the fact that $E\subset\bar{E}$ then gives \begin{align*} m^{\star,(J)}(E) &\leq m^{\star,(J)}(\bar{E})\\ &\leq\sum\limits_{j=1}^{N}|B'_{j}|\\ &\leq\sum\limits_{j=1}^{N}\left(|B_{j}|+\frac{\epsilon}{N}\right)\\ &\leq m^{\star, (J)}(E)+2\epsilon. \end{align*} In particular, we obtain the estimate $$\Bigg|m^{\star,(J)}(E)-m^{\star,(J)}(\bar{E})\Bigg|\leq2\epsilon,$$ and since $\epsilon$ was arbitrary, we conclude $m^{\star,(J)}(E)=m^{\star,(J)}(\bar{E})$ as required.

SOLUTION (2) By definition of $m_{\star, (J)}$ there exists boxes $\{B_{j}\}_{j=1}^{N}$ such that both \begin{align*} \bigcup\limits_{j=1}^{N}B_{j}\subset E &&\text{and} &&\sum\limits_{j=1}^{N}|B_{j}|\geq m_{\star,(J)}(E)-\epsilon \end{align*} hold for every $\epsilon>0$. Because $E\Delta E^{\circ}\subset\partial E\subset E'\cup\{x\in E:N_{\delta}(x)\cap E=\{x\}\forall\delta>0\}$, and isolated points have inner (and outer) measure $0$, by shrinking each box $B_{j}$ (as necessary), we can find a new collection of boxes $\{B'_{j}\}_{j=1}^{N}$ such each of the conditions \begin{align*} |B_{j}|\geq|B'_{j}|\geq|B_{j}|-\frac{\epsilon}{N}, &&\bigcup\limits_{j=1}^{N}B'_{j}\subset E^{\circ}\subset E &&\text{and} &&\sum\limits_{j=1}^{N}|B'_{j}|\geq m_{\star,(J)}(E^{\circ})-\epsilon \end{align*} also holds for the same $\epsilon$. Montonicity and the fact that $E^{\circ}\subset E$ then gives \begin{align*} m_{\star,(J)} &\geq m_{\star,(J)}(E^{\circ})\\ &\geq\sum\limits_{j=1}^{N}|B'_{j}|\\ &\geq\sum\limits_{j=1}^{N}\left(|B_{j}|-\frac{\epsilon}{N}\right)\\ &\geq m_{\star,(J)}(E)-2\epsilon. \end{align*} In particular, we obtain the estimate $$\Bigg|m_{\star,(J)}(E^{\circ})-m_{\star,(J)}(E)\Bigg|\leq2\epsilon,$$ and since $\epsilon>0$ was arbitrary, we conclude $m_{\star,(J)}(E)=m_{\star,(J)}(E^{\circ})$ as required.

SOLUTION (3) We have the inequality $$m_{\star,(J)}(E)=m_{\star,(J)}(E^{\circ})\leq m^{\star,(J)}(\bar{E})=m^{\star,(J)}(E).$$ In the case that $E$ is Jordan measurable, this immediately becomes an equality $$m_{\star,(J)}(E)=m_{\star,(J)}(E^{\circ})=m(E)=m^{\star,(J)}(\bar{E})=m^{\star,(J)}(E),$$ and monotonicity then implies that $m^{\star,(J)}(E\Delta\bar{E})=m^{\star,(J)}(E\Delta E^{\circ})=0$, which means that both $\bar{E}$ and $E^{\circ}$ are Jordan measurable with measure $m(E)$. Boolean closure then implies $\partial E$ is Jordan measurable since $\partial E=\bar{E}-E^{\circ}$, and from additivity we have $$m(\partial E)=m(\bar{E})-m(E^{\circ})=0.$$ Now suppose $\partial E=0$. Then sub-additivity of $m^{\star,(J)}$ and $m_{\star,(J)}$ implies $$0=m^{\star,(J)}_{\star,(J)}(\partial E)\geq m^{\star,(J)}_{\star,(J)}(\bar{E})-m^{\star,(J)}_{\star,(J)}(E^{\circ})\geq0,$$ so that the original inequality is extended to $$m_{\star,(J)}(E^{\circ})=m_{\star,(J)}(E)=m_{\star,(J)}(\bar{E})\leq m^{\star,(J)}(E^{\circ})=m^{\star,(J)}(E)=m^{\star,(J)}(\bar{E}).$$


(Update 1)

I want to look a little closer at my argument in (1). I can't see why it's incorrect, but perhaps a modification will make it more rigorous.

For convenience, let's used cubes $Q_{j}$ instead of boxes $B_{j}$ (that the former can be used in the definition of Jordan content is an easy consequence of the fact that any box $B_{j}$ can be arbitrarily approximated by a finite number of cubes by an obvious dissection process).

Then I have a collection of cubes $\{Q_{j}\}_{j=1}^{N}$ such that $\bigcup_{j=1}^{N} Q_{j}\supset E$ and $\sum_{j=1}^{N}|Q_{j}|\leq m^{\star,(J)}(E)+\epsilon$.

Suppose each cube has some index length $\ell_{j}$, and consider the cubes $\{Q_{j}'\}_{j=1}^{N}$ obtained by enlarging each $\ell_{j}\mapsto\ell_{j}+\frac{epsilon}{N}$.

Then this collection covers $\bar{E}$ for the following reasons. If $E$ has any isolated points, then the $Q_{j}$ already covered them, so the $Q'_{j}$ certainly cover them as well. Therefore, the only other points in $\bar{E}\backslash E$ are limit points of the set $E$. But limit points are arbitrarily close to $E$, the set covered by the $Q_{j}$ (they have distance $0$ from the cover), and since by the explicit construction $dist(Q_{j},Q'_{j})=\frac{\epsilon}{N}>0$ and $Q'_{j}$ also cover $E$, we conclude the $Q'_{j}$ also cover $\bar{E}$. (If this is wrong, pleeaaassseee explain to me the logical fault).

Then we have \begin{align*} m^{\star,(J)}(\bar{E}) &\leq\sum_{j=1}^{N}|Q'_{j}|\\ &=\sum_{j=1}^{N}\left(\ell_{j}+\frac{\epsilon}{N}\right)^{d}\\ &=\sum_{j=1}^{N}\sum_{k=0}^{d}\binom{d}{k}\left(\ell_{j}^{k}\left(\frac{\epsilon}{N}\right)^{d-k}\right)\\ &=\sum_{j=1}^{N}\left(\ell_{j}^{d}+d\frac{\epsilon}{N}\ell_{j}^{d-1}+\ldots+\frac{\epsilon^{d}}{N^{d}}\right)\\ &=\sum_{j=1}^{N}|Q_{j}|+NO\left(\frac{\epsilon}{N}\right)\\ &\leq m^{\star,(J)}(E)+O(\epsilon) \end{align*} which shows that the outer measure of $\bar{E}$ is less than the outer measure of $E$.

I still don't know if this is fully rigorous or not though. It still has the problem that the argument for outer Lebesgue measure carries over by using $\ell_{j}'=\ell_{j}+\frac{\epsilon}{2^{j}}$ as each of the terms in the binomial expansion now are convergent infinite series and the result is still $O(\epsilon)$. Moreover, we would still also have $dist(Q_{j},Q'_{j})=\frac{\epsilon}{2^{j}}>0$.

So I guess that basically proves there's something wrong with my logic. I just don't know how to fix it.

(Update 2)

So I have concluded there is nothing inherently wrong with my argument, just that it is incomplete. For one thing, you cannot conclude that an "$\epsilon$-close" countable covering of a set $E$ can be "$\epsilon$-fattened" to produce a cover of $\bar{E}$. The counter example is $[0,1]\cap\mathbb{Q}$. The reason for this somewhat peculiar fact is illustrated in this question I asked earlier today (Paradox as to Measure of Countable Dense Subsets?). But to put it summarily, countable coverings allow you to circumvent density arguments, in the sense that you can still cover $E$ while avoiding $\bar{E}$, and in fact do so on a set of positive outer measure; hence once $\epsilon$ is sufficiently small, the fattened cubes will still fail to cover $\bar{E}$. That this cannot happen in the finite case is made rigorous in the following (completely) redone solutions to (1) and (2). Actually, fattening of the original cover of $E$ is not even necessary, as it is shown that a finite cover of $E$ is necessarily a finite cover of $\bar{E}$, leading to a new characterization of outer Jordan measure involving closed boxes only. Moreover, we also obtain another surprising explanation as to why countable coverings can avoid this conclusion, the main idea being that countable unions of closed sets need not be closed. The same arguments apply in reverse for the case of inner Jordan measure of $E$ and $E^{\circ}$, where it is shown that any containment by $E$ is necessarily a containment by $E^{\circ}$, hence leading to a new definition of inner Jordan measure involving open boxes only. It is interesting to note that countable containments cannot avoid this conclusion (in contrast to the analogous situation for covers), and this stems from the fact that countable unions of open boxes are open. Incidentally, we conclude that nothing is gained by considering countable containments, hence why we do not consider a Lebesgue inner measure.

I will formulate this all into a complete answer once I prove (3).

Let $E\subset\mathbb{R}^{d}$ be a bounded set.

(1) By definition of $m^{\star,(J)}$, there exists a collection of boxes $\{B_{j}\}_{j=1}^{N}$ such that both \begin{align*} E\subset\bigcup\limits_{j=1}^{N}B_{j} &&\text{and} &&\sum\limits_{j=1}^{N}|B_{j}|\leq m^{\star,(J)}(E)+\epsilon \end{align*} hold for every $\epsilon>0$. Now the definition of elementary measure implies $m(B_{j})=|B_{j}|=\bar{B_{j}}|=m(B_{j})$, and because $\bigcup_{j=1}^{N}\bar{B_{j}}\supset E,$ we may assume each $B_{j}$ is closed. But this implies that the cover is itself closed since $N$ is finite, e.g. $\bigcup_{j=1}^{N}B_{j}=\overline{\bigcup_{j=1}^{N}B_{j}}.$ Moreover, as $\bar{E}$ is the \emph{smallest} closed set which contains $E$, we have that $\{B_{j}\}_{j=1}^{N}$ is an ``$\epsilon$-close'' cover of $\bar{E}$ as well. It follows from monotonicity and the previous remarks that $$m^{\star,(J)}(E)\leq m^{\star,(J)}(\bar{E})\leq\sum_{j=1}^{N}|B_{j}|\leq m^{\star,(J)}(E)+\epsilon\leq m^{\star,(J)}(\bar{E})+\epsilon.$$ In particular, we obtain the estimate $$\Bigg|m^{\star,(J)}(E)-m^{\star,(J)}(\bar{E})\Bigg|\leq\epsilon,$$ and since $\epsilon$ was arbitrary, we conclude $m^{\star,(J)}(E)=m^{\star,(J)}(\bar{E})$ as required.

(2) By definition of $m_{\star,(J)}$, there exists a collection of boxes $\{B_{j}\}_{j=1}^{N}$ such that both \begin{align*} E\supset\bigcup\limits_{j=1}^{N}B_{j} &&\text{and} &&\sum\limits_{j=1}^{N}|B_{j}|\geq m_{\star,(J)}(E)-\epsilon \end{align*} hold for every $\epsilon>0$. Now the definition of elementary measure implies $m(B_{j})=|B_{j}|=|B_{j}^{\circ}|=m(B_{j}^{\circ})$, and because $E\supset\bigcup_{j=1}^{N}B^{\circ}_{j}$, we may assume each $B_{j}$ is open. But this implies that the contained set is itself open, e.g. $\bigcup_{j=1}^{N}B_{j}=\left(\bigcup_{j=1}^{N}B_{j}\right)^{\circ}.$ Moreover, as $E^{\circ}$ is the \emph{largest} open set which is contained in $E$, we have that $\{B_{j}\}_{j=1}^{N}$ is an ``$\epsilon$-close'' set contained in $E^{\circ}$ as well. It follows from monotonicity and the previous remarks that $$m_{\star,(J)}(E)\geq m_{\star,(J)}(E^{\circ})\geq\sum_{j=1}^{N}|B_{j}|\geq m_{\star,(J)}(E)-\epsilon\geq m_{\star,(J)}(E^{\circ})-\epsilon.$$ In particular, we obtain the estimate $$\Bigg|m_{\star,(J)}(E)-m_{\star,(J)}(E^{\circ})\Bigg|\leq\epsilon,$$ and since $\epsilon$ was arbitrary, we conclude $m_{\star,(J)}(E)=m_{\star,(J)}(E^{\circ})$ as required.

(Some Remarks) These results allow us to redefine $m^{\star,(J)}$ as the infimal elementary measure of a cover of $E$ by a finite number of closed boxes, and $m_{\star,(J)}$ as the supremal elementary measure of a containment by $E$ consisting of finite numbers of open boxes. These characterizations are are just as valid for countable coverings and containments. However, to obtain the conclusion from (1), finiteness of the cover is essential. The issue is that a countable union of closed boxes need not be closed, and so we can not conclude that the outer measures of $E$ and $\bar{E}$ coincide when countable covers are permitted. The set $A=[0,1]\cap\mathbb{Q}$ is a typical example of how the conclusion of (1) can fail when countable coverings are allowed. The Lebesgue outer measure of $A$ is $0$ since it can be covered by a countable union of degenerate boxes, while the Jordan outer measure of $A$ is $1$ since $\bar{A}=[0,1]$. On the other hand, no problem occurs when finite containments are upgraded to countable containments. This is because a countable union of open boxes is always open, and so the "Lebesgue inner measure" of $E$ must still coincide with the Lebesgue inner measure of $E^{\circ}$, since $E^{\circ}$ is the largest open set contained in $E$. Moreover, every open set $\mathscr{O}$ is the countable union of pairwise disjoint boxes, and by being a union of such boxes (as opposed to an intersection), we see that every finite subcollection of such boxes is contained in $\mathscr{O}$. It is easy to see then that the inner Jordan and Lebesgue measures of $\mathscr{O}$ coincide, and the conclusion from (2) implies that they agree for all sets. Incidentally, this demonstrates our lack of need for an inner Lebesgue measure. In fact, Littlewood's characterization of Lebesgue measurable sets being ``nearly open'' is basically equivalent to showing that the inner and outer Lebesgue measures are equal; in any case, clearly nothing is gained by upgrading finite containments to countable ones.

  • 2
    $\begingroup$ I found reading all this very useful. However, couldn't the last update be posted as an answer? $\endgroup$
    – kekkonen
    Jul 7, 2014 at 16:58
  • 1
    $\begingroup$ I just don't understand why you complicate things so much. Note that elementary sets are finite union of boxes. The measure for boxes are always the same no matter whether they are open boxes or closed boxes. A collection of closed boxes that cover $E$ should definitely cover $\bar E$. A collection of open boxes that is contained by $E$ is definitely contained by $E^\circ$. $\endgroup$
    – Troy Woo
    Sep 27, 2014 at 19:47
  • 1
    $\begingroup$ Your initial proof of (3) stops short of making all quantities equal (there is an inequality in the middle). I think you still need to prove that the outer and inner Jordan measures of E are equal in order to prove Jordan measurability. $\endgroup$
    – Kevin
    Aug 26, 2020 at 1:24
  • 1
    $\begingroup$ I think you should separate your question into to both a question and and answer. $\endgroup$
    – Kevin
    Aug 26, 2020 at 1:25
  • $\begingroup$ A proof of (3) is outlined on page 4 here: wiki.math.ntnu.no/_media/tma4225/2015h/tma4225-f15-homework.pdf $\endgroup$ Aug 31, 2021 at 6:06

1 Answer 1


The following proof is intuitive enough to be summarized in a brief sketch:

  • (i) A closed elementary set containing $E$ necessarily contains $\bar E$
  • (ii) An open elementary set contained in $E$ is necessarily contained in $E^\circ$
  • (iii) On the one hand, find elementary set $A,B$ such that $A\subset E\subset B$ and $B\backslash A$ serves as an elementary set containing $\partial E$; on the other hand, given an elementary set $C$ containing $\partial E$, find the corresponding elementary sets $A,B$.

${\bf Proof:}$

${\bf (i)}$ By definition of outer measure, we can find an elementary set $B$ containing $E$. $B$ is by definition, the disjoint union of $N$ boxes $B_i$'s. It is apparent that $|B_i|=|B_i^\circ|=|\bar B_i|$. Therefore, $\bar B=\cup_{i=1}^N\bar B_i$ has the same Jordan measure as $B$. Also $\bar E\subset\bar B$. Therefore: $$ m^{\star,(J)}(\bar E)\leq m(\bar B)=m(B) $$ Take the infimum over all $B\in\varepsilon(\mathbb R^d)$ and we have: $$ m^{\star,(J)}(\bar E)\leq m^{\star,(J)}(E) $$ On the other hand, since $E\subset\bar E$, we have: $$ m^{\star,(J)}(E)\leq m^{\star,(J)}(\bar E) $$ and therefore $m^{\star,(J)}(E)=m^{\star,(J)}(\bar E)$;

${\bf (ii)}$ Similar to ${\bf (i)}$;

${\bf (iii)}$ If we have $E$ is Jordan measurable, there exists elementary sets $A,B$ such that $A\subset E\subset B$ and $m(B\backslash A)\leq\varepsilon>0$ for arbitrary small positive number $\varepsilon>0$. Recall $A,B$ is the disjoint union of finitely many boxes: $$ A=\cup_i B_{A,i},\qquad B=\cup_j B_{B,j} $$ Obviously: $$ \underbrace{\cup_i B_{A,i}^\circ}_{A^\circ}\subset\cup_i B_{A,i}\subset E\subset\cup_j B_{B,j}\subset\underbrace{\cup_j\bar B_{B,j}}_{\bar B} $$ w.l.o.g, we denote the two new elementary sets by $A^\circ\subset A$ and $\bar B\supset B$ to indicate they are open and closed respectively. It is easy to see that: $$ \bar E\subset\bar B $$ and $$ A^\circ\subset E\quad\Rightarrow\quad E^c\subset (A^\circ)^c\quad\Rightarrow\quad \overline{E^c}\subset (A^\circ)^c $$ since $(A^\circ)^c$ is closed. Therefore, $$ \partial E=\bar E\cap\overline{E^c}\subset \bar B\cap(A^\circ)^c =\bar B\backslash A^\circ $$ $\partial E$ has outer measure zero since $$ m(\bar B\backslash A^\circ)=m(B\backslash A)\leq\varepsilon $$ can be arbitrarily small.

Next, assume we already have an elementary set $C$ with arbitrarily small measure and $\partial E\subset C$. Now by Boolean closure, $C^c$ is also elementary, which means $C^c$ is the disjoint union of finitely many boxes $B_k$'s. $$ C^c=\cup_kB_k $$

We claim that: $$ \text{either }B_k\subset E\text{ or }B_k\subset E^c $$ For otherwise, exists $x\in E, y\in E^c$ such that $x,y\in B_k$ for some $B_k$. Since $B_k$ is convex, the line segment $\overline{xy}$ is contained in $B_k$ too. Construct the function: $$ \gamma:[0,1]\to B_k\subset\mathbb R^d, \gamma(0)=x, \gamma(1)=y,\gamma(t)=x+t(y-x) $$ which is obviously continuous. Now look at the preimage of $\bar E$ and $\overline{E^c}$. We have: $$ \gamma^{-1}(\bar E)\cup\gamma^{-1}(\overline{E^c})=\gamma^{-1}(\bar E\cup\overline{E^c})=\gamma^{-1}(\mathbb R^d)=[0,1] $$ and $$ \gamma^{-1}(\bar E)\cap\gamma^{-1}(\overline{E^c})=\gamma^{-1}(\partial E)=\varnothing $$ since $B_k\subset C^c$ and $C^c\cap\partial E=\varnothing$. We have $[0,1]$ as the disjoint union of two non-empty (why?) closed sets $\gamma^{-1}(\bar E)$ and $\gamma^{-1}(\overline{E^c})$, which contradicts the fact that $[0,1]$ is connected.

Now we are allowed to pick out all $B_{k}$'s in $C^c$ such that $B_{k}\subset E$ (denoted $B_{i,s}$'s) and simply discard all $B_{k}$'s outside $E$ (denoted $B_{o,s}$'s). Take the union of such boxes and call it $A$. To find $B$, simply take the disjoint union of $A$ with $C$: $B=A\cup C$. It covers $E$ since all the $B_i$'s we discarded are completely contained in $E^c$. Obviously $C=B\backslash A$. And we are done.

${\bf Remark:}$ The proof of ${\bf (iii)}$ is apparently motivated by a pictorial understanding...I am an engineer and this is what I do...

Edit: A technical mistake in the proof. One should not take the complement of $C$ w.r.t. $\mathbb R^d$, but w.r.t. a closed and bounded box containing $C$, for otherwise, the jordan measure is not well defined. I could no longer change the gif (damn I lost my .AI)...

Here an animated gif I made today for your consideration. enter image description here

  • $\begingroup$ How did you make this nice gif? :) $\endgroup$
    – AfterMath
    Aug 30, 2019 at 10:58
  • 1
    $\begingroup$ @AfterMath trade secret $\endgroup$
    – Troy Woo
    Sep 3, 2019 at 14:48
  • $\begingroup$ I don't think you can use the set complement as you have done here. A complement of an elementary set is not elementary. Elementary sets are finite unions of "boxes", and "boxes" are the $R^d$ equivalent of bounded intervals, such as $(a, b]$. A complement of an elementary set would require either countably many boxes, or finitely many unbounded boxes. $\endgroup$
    – Kevin
    Aug 26, 2020 at 1:19
  • $\begingroup$ I don't think that interior of the union is the union of interiors. Take, for example, a square, divide it into four pieces. The inner boundaries are in fact in the interior of the big square. What is true, however, is $\begin{aligned}A&=B\cup C\\\implies \operatorname{Int}B\cup\operatorname{Int}C&\subseteq\operatorname{Int} A.\end{aligned}$ $\endgroup$
    – PinkyWay
    Mar 16 at 7:12
  • $\begingroup$ As for the claim that the interior of an elementary set is elementary itself, I think, $\mathcal E$ being a ring, and in turn closed under set differences, guarantees you could just take out the outer edges in your disjoint union for $A\in\mathcal E.$ There are finitely many such edges, of course, maybe that might work out. $\endgroup$
    – PinkyWay
    Mar 16 at 7:22

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