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I am self teaching myself so I couldn't ask any teacher but here are somethings in Chapter 1 that I can't understand. In the book Area is defined axiomatically but some parts of it are not just making sense.

We assume there exists a class $M$ of measurable sets in the plane and a set function whose domain is $M$. - Chapter 1,page 58

First, what does he mean by sets in a plane? Does the set themselves hold some geometric properties? Because till now I have always encountered sets having numbers and ordered pairs as elements.

And in the axiom regarding congruence of sets in plane(?), he defines congruence as one-one correspondence between points (elements?) of set such that distances are preserved.

Again what does he mean by distance and how does that make sense if a set have just numbers as elements?

These are probably very poor question but I would really like some help. Thanks

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  • $\begingroup$ "Sets in the plane"= subsets of $\;\Bbb R^2\;$ , which of course are sets whose elements are ordered pairs. Perhaps you should mull over studying a good introductory book/course in set theory before attempting to study these things. $\endgroup$ – DonAntonio May 31 '16 at 8:40
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We assume there exists a class $M$ of measurable sets in the plane and a set function whose domain is $M$

Here, every member of $M$ is a subset of the plane $\Bbb R^2$, the set of ordered pairs of real numbers. Every $X\in M$ is a set of ordered pairs of reals: $X\subseteq \Bbb R^2$.

Disclaimer: I don't have a copy of Apostol's classic at hand.

If he doesn't define "measurable" by page 58, I'm sure he does somewhere, perhaps in an appendix. At very least you should assume that $M$ contains all basic geometric shapes (in all rotations and orientations, at all possible offsets): anything you intuitively thing of as having an "area". $M$ will also contain all finite unions of such shapes (and more still, but this is enough to think about now).

The measure $m(X)$ of a measurable set $X\in M$ is, roughly speaking, its area: whenever $X$ is a shape that has an area in the usual/intuitive sense, $m(X)$ will be that number, a nonnegative real. Thus, measure is a function $m\colon M \to \Bbb R_{\ge 0}$.

I expect that $M$ contains still more subsets of the plane than those mentioned, for example many countable unions of basic shapes. The measure of a more general "measurable set" is defined by approximating the set with unions of more basic shapes and summing their areas.


You relate that Apostol defines a congruence to be a 1-1 function of the underlying space — the plane $\Bbb R^2$, for example, or any $\Bbb R^n$ — which preserves the distance between points. "distance" is just that: the Euclidean distance you know, such as $$d((x,y), (u,v)) = \sqrt{(x-u)^2 + (y-v)^2} $$ in $\Bbb R^2$. In $\Bbb R$ (i.e. $\Bbb R^1$), the distance between reals $x$ and $u$ is just $\lvert x - u\rvert = \sqrt{(x-u)^2}$.

Thus, $\, f\colon \Bbb R^2 \to \Bbb R^2$ is a congruence iff $f$ is 1-to-1 and $$d(f(x,y), f(u,v)) = d((x,y), (u,v))\qquad\text{for all $(x,y), (u,v) \in \Bbb R^2$}. $$

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  • $\begingroup$ Thank you. This answers a lot but I still have one questions left. You say $M$ contains all basic geometric shapes if plotted on a $X-Y plane$ but I can't figure out how a finite set of ordered pairs can form a closed figure with an area as any finite set of ordered reals can always be represented by straight lines. So all of $X\subseteq \mathbb{R}^{2}$ which are measurable must have infinite elements. Am I correct? $\endgroup$ – user127 May 31 '16 at 12:02
  • $\begingroup$ That's right: the elements of $M$ are not finite, except for "degenerate cases". Each contains not just the the boundary of a "basic shape" but also the interior of the shape. (A set consisting of a single point, a singleton $\{\bf p\}$, is "degenerate": it's a square whose lower left corner is $\bf p$ and whose sides have length $0$; its area is $0$. Any finite set of points is a finite unions of singletons, and its area/measure is also $0$.) $\endgroup$ – BrianO May 31 '16 at 13:56

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