Conceptualizing the concept of outer measure.

Question

I'm trying to understand the definition of outer measure from a conceptual point of view. The definition is: if $A$ is an arbitrary subset of $\mathbb{R}$ its outer measure $m^*(A)$ is defined as:

$$m^*(A) = \inf\{\sum_i \ell (I_i): A \subset \bigcup_{i=1}^\infty I_i\}$$

Where $\bigcup_{i=1}^\infty I_i$ is a countable collection of intervals $(a_i, b_i)$ and $\ell (a,b) = b-a$.

I can conceptualize this definition in $\mathbb{R}$ but am having trouble conceptualizing in other spaces. In $\mathbb{R}^1$ I understand the definition to be the smallest countable open covering for the set. However if the set is in $\mathbb{R}^2$ how do we create a cover for a two dimensional set out of intervals? Ahh, maybe I just answered my own question. Is outer measure only defined for $\mathbb{R}$ and not for $\mathbb{R}^n$?

My apologies if this seems like such a trivial question. I have found that if I don't ask these types of questions I will never clear up my misconceptions.

Thank you.

Answer 1

In $\mathbb R^2$, the "intervals" become "squares". More generally, in $\mathbb R^n$ you can take "boxes" $$ B=\{(x_1\ldots,x_n):\ a_j\leq x_j\leq b_j:\ j=1,\ldots,n\}, $$ and $$ \ell(B)=(b_1-a_1)\cdots(b_n-a_n). $$

Answer 2

In a topological space, your basic objects are open sets. To define a measure that plays nice with the topology, you have to define that measure on the Borel sets. The usual way to do this is to define the measure on some basis for the open sets, then use that to figure out how to provide some notion of measure (specifically, an outer measure) to any set. Restrict to the measurable sets (a la Caratheodory) and you done.

In $\mathbb{R}$, the open sets are generated by open intervals, hence it makes sense to build a measure by first defining it on the intervals. Moreover, intervals have a very natural notion of measure already—their length. Working up to $\mathbb{R}^2$, a rectangle has a very natural notion of measure (its area), and rectangles of the form $(a,b)\times (c,d)$ generated the open sets. Hence it is reasonable to try to get a measure from that. Higher dimensional boxes give higher dimensional measures.

The point is that "intervals" are not part of the definition of the outer measure in a more general space. "Intervals" are specific to $\mathbb{R}$. If you want to define a Borel measure on some other space, you have to replace "interval" with "basis open set."

Answer 3

I understand the definition to be the smallest countable open covering for the set.

It's not the smallest, but the limit. Consider the closed interval [0,1]. There is no "smallest" open covering of it. You can take a convering of (0-$\epsilon$,1+$\epsilon$), which has a measure of 1+2*$\epsilon$. Taking the limit as $\epsilon$ -> 0, the outer measure is 1. For two dimensions, one way of visualizing the outer measure of, say, a circle, is to imagine drawing a shape around the circle using just horizontal and vertical lines. You'll have to leave some space between your shape and the circle, but that space can be made arbitrarily small. There are then various arguments that you can use to find that the outer measure is $\pi$r².

Now, of course, if you're using the open ball topology, you can just cover the circle with a circle. But the open rectangle topology is probably easier to visualize, since every "nice" shape can be covered with countably many rectangles with arbitrarily small overlap, and the area of a cover can be found by just taking the sum of length*width. An open triangle/simplex topology is also nice, since every polygon can be decomposed into triangles.