The Sample Space for a Die

Also featuring: a non-Lebesgue measurable set and Dynkin’s Lemma.

At the National Maths Summer School last week, the senior students and I spent a while talking about probability space, and in particular, when it was reasonable to assign a probability to a potential event. We considered rolling a standard die, and the probabilities $\mathbb{P}(D\in\{\})$ , the empty event, and $\mathbb{P}(D=7)$ . Though it is tempting to conclude that the latter must be zero, in the end we decided that it should not actually be defined at all.

Why? Well, if we accept $\mathbb{P}(D=7)=0$ , then by extension we must accept $\mathbb{P}(D=137)$ and $\mathbb{P}(D=\$1.50)$ also both exist and are zero. What have we gained? In reality very little. But the cost is this: we might define an event to be any subset of the sample space. Before, our sample space was $\Omega=\{1,2,3,4,5,6\}$ , and so there are exactly 64 events, including the possibly counter-intuitive empty event $\{\}$ . This is finite, which is always nice. With the extra events, however, we must extend the sample space to $\Omega=\{1,2,3,4,5,6,7,\ldots$ , where “…” means “the rest of the universe”. This is a fairly exotic mathematical object, and really has no place in any sensible discussion.

This reminded me of one of my favourite results from Part II Probability and Measure. Of course, for uncountable sample spaces, we cannot necessarily assume all subsets of $\Omega$ are measurable. Instead we build up a sigma-algebra of measurable sets, most importantly for Lebesgue measure on $\mathbb{R}$ . An immediate question to ask is: are all subsets of $\mathbb{R}$ Lebesgue-measurable?

And the answer is ‘no’. Why? The standard counterexample is as follows. Consider Lebesgue measure on the unit interval $U=\mathbb{R}/\mathbb{Z}$ , with endpoints identified. Now consider the rationals in U, which are actually a subgroup $\mathbb{Q}\leq U$ , with uncountably many cosets. Pick an element from each coset (*). Call this sets A. Then, working modulo 1, $U=\cup_{q\in\mathbb{Q}\cap U}A+q$ . If A is Lebesgue-measurable, then so is A+q, and $\mu(A+q)=\mu(A)$ (**). Combining these two results, using countable additivity:

$\mu(U)=$ (0 if $\mu(A)=0, \infty$ otherwise).

This is a contradiction, and hence we conclude that A is not Lebesgue-measurable.

Remark on (*): This relies on the Axiom of Choice. In fact, the existence of non-Lebesgue measurable sets MAY be equivalent to AC.

Remark on (**): I was suddenly unsure that this was obvious. I mean, this is such a weird set that it is in fact not measurable: why should its hypothetical measure be translation invariant? It is tempting to argue vaguely, by saying that the construction of Lebesgue measure is invariant under translation at all steps. As so often with elementary measure theory, recourse to Dynkin’s Lemma is more reliable.

Let D be the collection of measurable sets whose measure is invariant under translation. By definition, D is invariant under translation (of its elements). D certainly contains all intervals in U, which is a pi-system generating $\mathcal{B}([0,1])$ . But, in a classic proof by suggestive notation, we can check that D is a d-system. The presence of the empty sets is clear. If we have $B\subset A$ , both in D, then also $A\backslash B\in D$ , as the translates of $x\in A\backslash B$ must be in A, but cannot be in B, as B is translation invariant. Finally, given $A_1\subset A_2\subset\ldots\subset D$ , then $x\in\cup A_i\Rightarrow x\in A_n$ for some n, so all of x’s translates are in $A_n$ , and hence in $\cup A_i$ .