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Supposing $\Omega = \mathbb{R}$ and $\mathcal{F} = 2^\Omega$. Then, I have read somewhere that it is not possible to define a set function $P:\mathcal{F} \to [0,1]$ such that $P$ satisfies $P(\cup_{i=0}^\infty A_i) = \sum_{i=0}^\infty P(A_i)$ whenever $A_i$ are pairwise disjoint and $P(\Omega) = 1$. Can somebody explain or point me to a good reference which proves this.

Any help is much appreciated.

EDIT: As ArturoMagidin pointed out, it is indeed possible to define such a $P$. I was trying to understand the need for defining a $\sigma$-algebra of sets and defining $P$ on those sets. I might have to put more conditions on $P$ but I dont know which ones.

Thanks, Phanindra

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What you state is not accurate: you can define $P\colon\mathcal{F}\to[0,1]$ by $P(A) = 0$ if $1\notin A$, and $P(A) = 1$ if $1\in A$. That function satisfies that $P(\Omega)=1$, and $P(\cup_{i=1}^{\infty}A_i) = \sum P(A_i)$ when the $A_i$ are pairwise disjoint, since at most one of them will contain $1$. What you read probably had a few more requirements on $P$. – Arturo Magidin Oct 12 '11 at 7:06
@ArturoMagidin: True. Thanks for pointing this out. I was trying to understand the need for defining a $\sigma$-algebra of sets (such as the Borel $\sigma$-algebra) and defining $P$ only on those sets. I might be missing a few more restrictions on $P$ as you have suggested. – jpv Oct 12 '11 at 7:14
You want a "measure" on $\mathbb{R}$ that satisfies the following conditions: (i) It is translation invariant. (ii) It is $\sigma$-additive. (iii) The "total measure" is finite nonzero. If you assume the Axiom of Choice, this is impossible: define an equivalence relation $x\sim y$ if and only if $x-y\in\mathbb{Q}$, and let $V$ be a set that contains one and only one representative from each equivalence class. Let $\{q_n\}$ be an enumeration of the rationals. Then $V+q_n\cap V+q_m=\emptyset$ if $n\neq m$, $\cup V+q_n=\mathbb{R}$. So $P(V)$ must satisfy $\sum_{i=1}^{\infty}P(V)=\mu(R)$. – Arturo Magidin Oct 12 '11 at 7:20
See Wikipedia's article on Vitali sets, and non-measurable sets. The "hard measure problem" usually requires that every subset be measurable, that every subset that has a "natural measure" (intervals, cubes, etc) have measure equal to that 'natural measure' (in $\mathbb{R}$, you want intervals $(a,b)$ to have measure $b-a$); that it be translation invariant; and that it be $\sigma$-additive. Assuming AC, the problem is unsolvable for $\mathbb{R}$. – Arturo Magidin Oct 12 '11 at 7:26
Probability measures aren't necessarily translation-invariant, though (nor can they be, if you want all of $\mathbb{R}$ to have measure $1$) – user7530 Oct 12 '11 at 9:58
up vote 2 down vote accepted

The "Hard measure problem on $\mathbb{R}$" is to find a nonnegative set theoretic function $\rho$ whose domain are subsets of $\mathbb{R}$, such that:

  1. $\rho(E)$ is defined for all $E\subseteq\mathbb{R}$;
  2. $\rho(I) = \mathrm{length}(I)$ for all intervals $I\subseteq\mathbb{R}$;
  3. $\rho$ is $\sigma$-additive;
  4. $\rho$ is invariant under isometries; i.e., if $j\colon\mathbb{R}\to\mathbb{R}$ is an isometry, then $\rho(j(E)) = \rho(E)$ for all $E\subseteq $\mathbb{R}$.

If we assume the Axiom of Choice, then the hard measure problem on $\mathbb{R}$ cannot be solved, as shown by the construction of Vitali sets. On the other hand, Solovay proved that in there is a model of Set Theory without the Axiom of Choice in which the hard problem can be solved.

The "Easy measure problem on $\mathbb{R}$" requires a measure that satisfies 1, 2, and 4 above, and replaced 3 with simple finite additivity. Banach proved that the problem can be solved (non-uniquely) in $\mathbb{R}$, and in $\mathbb{R}^2$ (where 2 replaces interval and length with rectangle and area), and Hausdorff proved that the easy measure problem cannot be solved in $\mathbb{R}^n$ with $n\geq 3$.

The does not directly address your question, because a measure that satisfies 2 above cannot have $\rho(\mathbb{R})=1$. H

A probability measure is just a function that takes values on $[0,1]$, is $\sigma$-additive, and satisfies $\rho(\emptyset)=0$ and $\rho(\mathbb{R})=1$. There are plenty of probability measures on all of $\mathbb{R}$ that are defined on every subset: you can define atomic measures in which there are countably many atoms and their measures add up to $1$, for instance. However, usually you want some sort of "uniformity" to the measure (e.g., there is no uniform probability measure on $\mathbb{N}$, or on any countable set). So one needs to specify exactly what conditions you want to place on the measure in order to determine whether or not one can construct such a probability measure.

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