# Is the exclusion of uncountable additivity a drawback of Lebesgue measure?

A friend and I were having a discussion about Lebesgue measure. I attempted to be profound by making the following points:

• Analytic geometry has been a fantastic tool, but the concept of representing a continuous "object" as a collection of points is inherently contrived (with a negative connotation). Immediately we run into the paradox that a point has no volume, and yet a collection of many points has volume.
• The notion of Lebesgue measure attempts to resolve this essential tension by allowing only countable additivity of the measure. But it does so only by disallowing certain operations (uncountable sums) that intuitively seem reasonable. As such, it is an indispensable tool, but it remains contrived on some level.

My friend countered by saying that uncountable additivity doesn't really make sense anyway, since any uncountable sum that converges must have co-countably many terms zero.

I would say I am still on the fence about this discussion. He makes a good point, but after all, it is exactly the addition of uncountably many zeros that I am concerned with, so the notion that co-countably many of the terms must be zero may not be a decisive objection.

• I am especially interested in @Qiaochuyuan 's opinion on this. – Eric Auld Jun 14 '14 at 17:45
• Please write down a proposed definition of "uncountable additivity". – Hans Engler Jun 14 '14 at 19:00
• @HansEngler One possible notion of an uncountable sum is: Suppose $\{x_i\}_{i\in I}$ is a collection of elements of a normed vector space. Then we say $\sum_{i \in I}x_i = x$ if for all $\epsilon >0$, there exists a finite subset $S_{\epsilon} \subset I$ such that for all finite subsets $T \subset I$ such that $T \supset S_{\epsilon}$, we have $|\sum_{i \in T}x_i - x|< \epsilon$. This is just the convergence of a net defined on the directed set $\mathcal{P}_0(I)$ of finite subsets of $I$, with join being union. Then uncountable additivity means $\mu(\cup_{j \in J}S_j) = \sum_{j \in J}\mu(S_j)$ – Eric Auld Jun 14 '14 at 19:17
• Some people feel that the right version of the additivity axiom should be that, as long as $\kappa<|\mathbb R|$, the union of $\kappa$ many disjoint measurable subsets of $\mathbb R$ should be measurable and have as measure the sum of the measures of the sets. (For any set $I$, countable or not, and for nonnegative reals $r_i$, $i\in I$, we define $\sum_{i\in I}r_i$ as the supremum of the sums of finitely many of the $r_i$.) (Cont.) – Andrés E. Caicedo Nov 18 '14 at 17:35
• This version of the additivity axiom is actually independent of the usual axioms of set theory It follows from the continuum hypothesis (trivially), from Martin's axiom, and from many other proposed independent statements of interest, while it is false under a variety of other such assumptions. – Andrés E. Caicedo Nov 18 '14 at 17:35

We have $[0,1] = \bigcup_{x \in [0,1]} \{ x \}$. How could you make a definition of uncounteable additivity which would produce $1 = \sum_{x \in [0,1]} 0$?