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if we start axioming area to be a mapping from a collection of points on a plane to the set of real numbers, this is: a: M → R S↦a(S), where S is the set of points as aforementioned and M is the collection of sets S (measurable sets..) . Stating the following axioms:

(i) a(S) is non negative for every S

(ii) if S and T are elements of M then S∪T and S∩T are also member of M and: a(S∪T) = a(S) + a(T) - a(S∩T)

(iii) Every retangule R is member of M and if h and k are sides of R then a(R)= k.h

--> Finaly, with thiss if we want to measure a area of anything, namely a retangule, we need to define the set of all point that compose this thing, so for example if we integrate the area of retangule with sides h and k we define the set

                A={(x,y):(0≤x≤h ^ 0≤y≤k)} what by the axioms: a(A)=h.k

but if we think that A as union of lines horizontal or vertical for example A being the reunion of all lines C={(x,t):0≤x≤h} where t obey 0≤t≤k and is diferent for each line, but thnking in this way lead us that a(A) is equal to the sum of all areas a(C) for all lines that compose A, but each area a(C) is equal to zero with implies that a(A)=0 what is clearly absurd!

My question is where is the failure in this procedure?

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  • $\begingroup$ Your axioms don't tell what happens when we have an infinite number of measurable sets and we take their union. The example which are taking involves such a procedure. $\endgroup$
    – Paramanand Singh
    May 13, 2016 at 6:14

1 Answer 1

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This is a good question! What you're getting at is the additivity of area. We certainly want the area of a (disjoint) union of two regions to be the sum of the areas of the regions, but what about infinite unions? What about uncountable unions?

If we want area to be totally additive - that is, the area of an arbitrary disjoint union $\bigsqcup_{i\in I} A_i$ is the sum of the areas, $\sum_{i\in I} m(A_i)$ - then you're right: if we want points to have zero area, this implies that every set has zero area. But we don't need area to be totally additive! In fact, the axiom you've written down only lets us prove finite additivity.

We usually do assume some infinite additivity - specifically, countable additivity: if $\{A_i: i\in\mathbb{N}\}$ is a countable collection of disjoint sets, $m(\bigcup A_i)$ is $\sum m(A_i)$. It turns out that this does not lead to everything having measure zero - this is basically an elaboration on Cantor's theorem that the set of real numbers is uncountable.

Now, you might ask: from a philosophical point of view, what justifies countable additivity as opposed to total additivity? I'm not going to get into that here, since it's a long story; but hopefully what I've written so far explains how we can have a rich theory of areas, without collapsing everything.

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  • $\begingroup$ Its great that you brought in various kinds of additivity. +1 $\endgroup$
    – Paramanand Singh
    May 13, 2016 at 6:15
  • $\begingroup$ thank you for the explanation. :) $\endgroup$ May 14, 2016 at 13:09

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