Problem For $D\subset\mathbb{R}^3$ be region such that

$D=\{(x,y,z)\in\mathbb{R}^3:0\leq x,y,z$ and $x+y+z\leq 1\}$.

Explain why $D$ is Jordan measurable, that is, show $1_B$ is Riemann Integrable.

Thoughts Intutively, this is a hexagon in $\mathbb{R}^3$ with finite volume, but I'm not sure if I understand Jordan measuribility. My definition here says that a set is Jordan measurable if $\varphi(bd(D))=0$, i.e. the outer measure of $bd(D)$. I'm having difficulty understanding outer measure and I don't have an intuition of what it is. Is there a visual intuition behind outer measure? What does it mean for the outer measure to equal zero? I've searched using google for more details, but I can only mostly find treatises or very lengthy papers written at a level beyond my understanding. Thanks in advance!

(outer measure is as its defined here http://en.wikipedia.org/wiki/Outer_measure)


Intuitively, the outer measure of a set $X$ in $\mathbb{R}^3$ is the greatest lower bound on the total volume of rectangles you'd need to "cover" the set. The idea is that you find a finite or countable infinite collection of three-dimensional rectangles which completely cover the set $X$. That is, each point in $X$ is contained in at least one of the rectangles.

The set $X$ can be covered by rectangles in many different ways: The outer measure of $X$ is defined as the greatest lower bound (infimum) of the set of all total volumes given by the collection. Each way $w$ of covering $X$ by rectangles produces a number $V_w$ which is the sum of the volumes of the rectangles in the cover. The outer measure is the greatest lower bound of the set $\{V_w\}$ of these numbers.

In other words. If you can cover $X$ with a collection of rectangles whose total volume is no more than $20$, then the outer measure is no more than than $20$. If you can cover the rectangle with some collection (which depends on $\epsilon$) of rectangles with total volume less than $10 + \epsilon$ for any $\epsilon > 0$, then the outer measure is no more than than $10 + \epsilon$. Since you could do this for any $\epsilon > 0$, then the outer measure is no more than $10$. If in addition it is impossible to cover $X$ with a collection of rectangles whose total volume is less than $10$, then the outer measure is $10$.

The outer measure of a set depends on the ambient space in which it exists. For example, the outer measure of a line segment considered as a subset of $\mathbb{R}^1$ is just the usual length of the segment. But the measure of a line segment (say a segment on the $x$-axis with $y = 0$ and $z = 0$) in $\mathbb{R}^2$ or $\mathbb{R}^3$ is zero. This is because, although the cover using rectangles has to cover the length of the line segment, the width and height of the rectangles can be made arbitrarily small.

According to Spivak's Calculus on Manifolds, a set is defined as Jordan-measurable if it is bounded and its boundary has measure zero.

Another way of looking at it is to recall that a function on a closed rectangle in $\mathbb{R}^n$ is Riemann integrable if and only if the set of discontinuities of the function is measure zero. In this case we're looking at the indicator function for the set $D$ in question, and this function is discontinuous on the boundary of the set.

The set $D$ is a tetrahedral region in the first octant (where all coordinates are at least zero) of $\mathbb{R}^3$ bounded by the $(x,y)$ plane, the $(y,z)$ plane, the $(x,z)$ plane and the plane $x + y + z = 1$. The boundary of this object is the collection of points in $D$ where either A) At least one of the coordinates is equal to zero, or B) $x + y + z = 1$ a triangular surface with corners $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. You can cover each of the "sides" for A) with one 1 x 1 x $\epsilon$ rectangle for any $\epsilon$. You can cover the triangular surface section in B) with the region between $x + y + z = 1 - \delta$ and $x + y + z = 1 + \delta$ .

The covers for A) and B) cover the boundary of $D$, and since you could make the total volume of the cover arbitrarily small, you just showed the boundary has outer measure zero. This means the set $D$ is by definition Jordan-measurable.

  • $\begingroup$ Thanks so much for your detailed response! Definitely helps a lot to get your clarification especially with regards to the set I provided. $\endgroup$ – jontameguchi Jun 14 '13 at 2:09

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