How is a set of two dimensions measured by integrating an indicator function?

The indicator function $\chi$ of a set $A$ is that function for which

$$\chi_A(x) = \begin{cases} 1 & \text{if x\in A} \\ 0 & \text{otherwise} \end{cases}$$

(Some people prefer $\mathbf{1}_A$ as a symbol instead.)

When dealing with a universal set $\Bbb U$ (usually $\Bbb R$ I presume) the measure of the set $A\subset\Bbb U$ is, as I understand it,

$$m(A) = \int_\Bbb{U} \chi_A(x) \, dx$$

(where $x$ is simply a dummy variable).

Of course, sets of two dimensions have measures. For example, $m([4,6]\times[0,10])=20$, which I could tell you having taken simple geometry. However, in general terms, how is the measure of a two-dimensional set defined in terms of an indicator function and integration?

Naturally, the element of each set will be an ordered pair, and I anticipate that the measure of the two-dimensional set is not necessarily the product of the measures of each dimension.

• Based on what you've written, I'm not sure what you think is missing. Why not just take your general statement in the case $\mathbb{U}=\mathbb{R}^2$? – Eric Wofsey Oct 30 '17 at 2:25
• @EricWofsey I'm not sure how that would look. Wouldn't I need a double integral then? – gen-z ready to perish Oct 30 '17 at 2:26
• Well that depends on your definitions of integrals. Usually, people define measures first and then define integrals in terms of them, not the other way around. – Eric Wofsey Oct 30 '17 at 2:27
• @EricWofsey I'm trying to sort through the eclectic assortment of lecture notes over measure theory I can find while I look for a decent book, but I'm pretty lost ¯\_(ツ)_/¯ – gen-z ready to perish Oct 30 '17 at 2:29

The measure of a two-dimensional set can be defined exactly as you have written it -- if the characteristic, or indicator, function $\chi_A(x)$ is defined to be $1$ over the set $A$ and $0$ elsewhere, the measure of a set $A$ is indeed equal to $$m(A) = \int_\mathbb{U} \chi_A dm$$
This is the notation I learned in Rudin -- instead of integrating with respect to a bound variable x ($dx$) you integrate with respect to the measure $m$. Does this seem circular?
The thing is -- if I am remembering correctly -- we can resolve this with exactly the sort of simple set you bring up. The measure of a rectangle in $\mathbb{R}^2$, or a box in $\mathbb{R}^n$ (regardless of we include or exclude the vertices/edges/faces/etc.) is defined to be the product in $\mathbb{R}$ of the lengths of the sides. This is useful because every set that can be written as the union of a countable collection of disjoint boxes can be measured by breaking it into boxes, measuring those with the defined measure, and summing.