8
$\begingroup$

There exists no map $\mu: \mathcal{P}(\mathbb{R})\rightarrow [0,\infty]$ satisfying:

(1) $\mu ((a,b])=b-a$ for all $a,b\in\mathbb{R}$ (this is what I mean by non-trivial in the title),

(2) Translation invariance,

(3) Countable additivity.

We call one such map a measure on $\mathcal{P}(\mathbb{R})$. A reason for its non-existence lies in the properties of the Vitali set $V$, that imply that $V$ cannot belong in any $\sigma$-algebra on which a measure satisfying (1), (2) and (3) exists.

As a relaxation of (3) consider

(3)* Finite additivity.

My question is whether a map $\mu: \mathcal{P}(\mathbb{R})\rightarrow [0,\infty]$ satisfying (1), (2) and (3)* exists.

Let's call this a finitely additive measure on $\mathcal{P}(\mathbb{R})$. If one such map $\mu$ existed then the Vitali set $V$ would have to satisfy $\mu(V)=0$.

A non-exitence proof here could be something along the lines of the Banach Tarski Paradox, but in $\mathbb{R}$.

Bonus question: What about a map $\mu: \mathcal{P}(\mathbb{R})\rightarrow [0,\infty]$ satisfying (1) and (3)?

Finally, I wonder what's so great about countable additivity that someone decided that all measures should satisfy it.

Related results:

  • If two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system) and satisfy that $\mu_1(X)=\mu_2(X)<\infty$, then $\mu_1=\mu_2$ on the $\sigma$-algebra generated by $\mathcal{A}$ (by the $\pi-\lambda$ Theorem). For example two finite measures that agree on all open sets also agree on all Borel sets.

  • [Generalization of the above] Suppose two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system). Moreover suppose there exists a countable nested subfamily of $\mathcal{B}\subseteq \mathcal{A}$ of sets that cover $X$ and have finite $\mu_1$-(and $\mu_2$-) measure. Then $\mu_1=\mu_2$ in the $\sigma$-algebra generated by $\mathcal{A}$ (again using the $\pi - \lambda$ Theorem). So for example the Lebesgue measure in all Borel sets in $\mathbb{R}^n$ is uniquely determined by its values on boxes/intervals.

  • Up to a multiplicative constant, Lebesgue measure is the only translation-invariant measure on the Borel sets that puts finite measure on the unit interval. This can be generalised to higher dimensions considering the unit box $[0,1]^n$ instead of the unit interval. You can find proofs here using, you guessed it, the $\pi - \lambda$ Theorem. The proof using the equivalent Monotone Class Theorem also seems quite straightforward here.

  • Let $\mu_1$ and $\mu_2$ be $\sigma$-finite measures on a measurable space $(X,\mathcal{S})$, $\mathcal{A}$ an algebra which generates $\mathcal{S}$ and suppose that, for each $A\in\mathcal{A}$, $\mu_1(A)=\mu_2(A)$. Then $\mu_1(B)=\mu_2(B)$ for each $B\in\mathcal{S}$ (still haven't seen a proof for this).

In relation to the bonus question, by the second (or third) related result any measure $\mu$ satisfying (1) is equal to the Borel measure on all Borel sets. And also, if all Lebesgue measurable sets are $\mu$-measurable, equal to the Lebesgue measure on said sets, since the completion of a measure space is unique. So the question really has to do with whether the Lebesgue measure can be extended to a measure on all of $\mathcal{P}(\mathbb{R})$ (see t.b.'s comment here).

$\endgroup$
4
  • 1
    $\begingroup$ If I remember correctly, an application of Hahn-Banach theorem shows the existence of such a measure. $\endgroup$ Jul 6, 2017 at 10:48
  • $\begingroup$ I think I remember seeing the existence proof in Terence Tao's book. $\endgroup$ Jul 7, 2017 at 3:14
  • $\begingroup$ Related question. $\endgroup$
    – bof
    Jul 7, 2017 at 3:58
  • 1
    $\begingroup$ This is only tangentially related, but for functions into $[0,1] \subset [0, \infty]$, de Finetti agreed with you that countable additivity is not that important, and that only finite additivity should be required. arxiv.org/pdf/1206.4769.pdf This paper wwwf.imperial.ac.uk/~bin06/Papers/favcarev.pdf also seems to have a quote very relevant to your specific concern: "length and area – Lebesgue measure on the line and the plane – can be defined for all sets, ... need not involve non-measurable sets, if one works with finite additivity." The same isn't true for $n \ge 3$. $\endgroup$ Jul 9, 2017 at 8:10

1 Answer 1

2
$\begingroup$

I believe the comments cover up the question, but I can somehow explain what is the importance of countable additivity in measures.

First of all, you should think about the importance of finite additivity: Consider a space $X$ and try to think of a measure as a scale that weights some subsets of $X$. Put two disjoint sets on your scale (just like putting two foreign objects); it is only natural to expect from your scale to return as weight value the sum of the two weights.

It is also quite natural to expect lower continuity: If $E_1\subset E_2\subset\dots$ then $\mu(\cup_{j} E_j)=\lim_{j\to\infty}\mu(E_j)$ Think of putting an object on the scale and then another object too, then one more object too, on and on. Well, the expected weight is the -theoretically speaking- value of the objects that will be on the scale at time = infinity, which is, obviously the limit. These two properties are equivalent to countable continuity though!

This simile helps me a lot. A dirac measure for instance is like putting the whole weight on a point. The lebesgue measure is like smashing the weight into powder and uniformly spreading it all over your space. Hope this helps.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .