# Compactness of sigma-algebra for the $L^1$ metrics

Consider a probability space $(X,F,\mu)$, and the quotient $G$ of the sigma-algebra $F$ by its null sets. Endow $G$ with the metric $d(A,B) = \mu(A \triangle B)$. Is $(G,d)$ a compact metric space?

When $F$ is generated by a (countable) partition, the answer is yes. Is there a general argument to prove that $(G,d)$ is always compact?

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Take the classical Bernoulli scheme with the base $(1/2,1/2)$, and let $A_n$ be the set where the $n$-th coordinate is 1. Then $d(A_n,A_m)=1/2$ whenever $n\neq m$. Since all purely non-atomic Lebesgue spaces are isomorphic, the same is applicable to any purely-non-atomic Borel measure on a Polish space.

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And the punch-line: this shows it is not compact, because there is an infinite set with all distances $1/2$. – Gerald Edgar Mar 17 '13 at 22:43

This space is certainly not compact in general, unless I misunderstood your question.

For instance, assume the probability space in question is $[0,1]$ with Lebesgue measure. Then endow the isometry group with the pointwise convergence topology; it would be a compact group if the space were compact. The automorphism group of $([0,1],\lambda)$ (i.e measure-preserving bijections) , sits inside that group as a closed subgroup (it is exactly the space of isometries that fix $\emptyset$, but you do not need to know that to see that a limit of measure-preerving bijections is still a measure-preserving bijection) and is certainly very much noncompact: a very indirect way to see this is to note that there exist dense conjugacy classes (any aperiodic element has a dense conjugacy class by Rokhlin's lemma), whereas conjugacy classes would be closed if the group were compact.

Actually, this space is, in a sense, as far as possible from being compact while staying separable: endow it with a group structure by using symmetric difference; the the corresponding Polish group is extremely amenable, i.e every continuous action of this group on a compact space has a (global) fixed point. This is thus impossible for the space to be compact, obviously, but also to be locally compact.

(Note: I'm used to working with groups so this is what I used in my answer, you can probably remove them very easily...)

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The metric space $(G,d)$ is well treated in Dunford and Schwartz. Of course, it is not compact for atomless probability spaces because the interval $[0,1]$ is not compact in $L_1$. It is however complete which is sufficiently amazing. – Rabee Tourky Mar 17 '13 at 10:20
@Julien Thanks for this answer. To which probability measures on [0,1], apart from the Lebesgue measure, does this apply? – Did Mar 17 '13 at 12:36
@Rabee Thanks. What do you call "not compact in $L^1$"? Which part of Dunford and Schwartz? – Did Mar 17 '13 at 12:38
@Didier Piau: Dunford and Schwartz, Part I, Section III.7 has a paragraph The metric space $\Sigma(\mu)$ plus some exercises in III.9. It appears in the index "measure space, as a metric space". – Martin Mar 17 '13 at 13:44

The space you are asking about embeds in a natural way into the unit ball of $L^\infty$ and the latter has three complete metric structures, each of independent interest and each containing your space as a closed subset.
2. the $L^1$ norm. this induces the metric you are interested in and is not compact as noted above. It is of great historical interest since it was used by Saks in his proof of his version of the Vitali-Hahn-Saks theorm.
3. the weak $\ast$ topology. This IS compact and is metrisable under suitable countability conditions on the probability space. (I added this info since I reinterpreted your question as: is there a natural compact topology on this space?).