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Let $X=2^{\omega}$ (the space of one-way infinite binary sequences) and $\mathcal{P}(X)$ the space of Borel probability measures on $X$. A measure $\mu \in \mathcal{P}(X)$ is absolutely continuous with respect to $\nu \in \mathcal{P}(X)$, written $\mu \ll \nu$, iff for every $\epsilon>0$ there is a $\delta>0$ such that for every Borel $B$, if $\nu(B)<\delta$ then $\mu(B) < \epsilon$.

Question 1 Does it suffice to check only cylinder sets? I.e. is it the case that $\mu \ll \nu$ iff for every $\epsilon>0$ there is a $\delta>0$ such that for every finite binary string $\sigma$ if $\nu([\sigma])<\delta$ then $\mu([\sigma]) < \epsilon$, where and $[\sigma] = \{ x \in X: \sigma \text{ is an initial segment of } x\}$?

Question 2 Is there some countable collection $\mathcal{C}$ of Borel sets such that $\mu \ll \nu$ iff for every $\epsilon>0$ there is a $\delta>0$ such that for every $B \in \mathcal{C}$, if $\nu(B)<\delta$ then $\mu(B) < \epsilon$?

Of course, an affirmative answer to Question 1 implies an affirmative answer to Question 2.

Quinn Culver
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  • This does not really answer your question, but you might be interested to know that an equivalent definition is as follows: "$\mu\ll\nu$ if and only if $\mu(B)=0$ for every Borel $B$ such that $\nu(B)=0$." – M Turgeon Oct 02 '12 at 19:46
  • @MTurgeon Thanks, I'm aware of that definition. My goal is (or 'was' rather, now that Davide Giraudo has shown me the light) to avoid quantifying over an uncountable set (e.g. for every Borel $B$...), which is why I was considering the $\epsilon$-$\delta$ characterization. – Quinn Culver Oct 04 '12 at 12:11
  • @DavideGiraudo I think you should leave it since it at least answers Question 2. – Quinn Culver Feb 07 '13 at 03:11
  • @DavideGiraudo I have answered Question 1. Would you please (check my answer!) and repost your answer to Question 2. I will (re-)accept your answer. – Quinn Culver Feb 07 '13 at 20:27

2 Answers2

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Recall

Let $(\Omega,\mathcal B,m)$ a finite measure space, and $\mathcal A\subset\mathcal B$ a generating algebra. Then for each $\varepsilon>0$, we can find $A\in\mathcal A$ such that $m(A\Delta B)<\varepsilon$.

(see here for a proof). We shall apply this result to $\mathcal A$ the algebra generated by cylindrical sets.
Fix $\varepsilon>0$. We can find $\delta>0$ such that for each $A\in\mathcal A$ satisfying $\nu(A)\leqslant 2\delta$ then $\mu(A)\leqslant \varepsilon$. Let $B\in\mathcal B(X)$. We can find $A\in\mathcal A$ such that $\mu(A\Delta B)+\nu(A\Delta B)<\delta$. If $\nu(B)\leqslant \delta$, then $\nu(A)\leqslant |\nu(B)-\nu(A)|+\nu(B)\leqslant 2\delta$ hence $\mu(A)\leqslant \varepsilon$. We can assume that $\delta<\varepsilon$. Then $\mu(A)\leqslant |\mu(B)-\mu(A)|+\mu(B)\leqslant 2\varepsilon$.

So if for all $\varepsilon>0$, we have can find $\delta>0$ such that if $A\in\cal A$ satisfies $\nu(A)\leqslant 2\delta$ then $\mu(A)\leqslant \varepsilon$, the same holds replacing $\cal A$ by $\mathcal{B}(X)$.

The algebra generated by cylindrical subsets is countable.

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Davide Giraudo
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    Thanks. I think since $\mathcal{A}$ is the collection of finite unions of cylinders, your answer to 2. is technically incorrect. – Quinn Culver Oct 04 '12 at 12:07
  • @QuinnCulver Actually I only gave a hint for 2. (but I should have mentionned it explicitely). Now I think it's more correct. – Davide Giraudo Oct 04 '12 at 14:50
  • So your answer doesn't actually imply that the answer to my Question 1. is "yes" since I only ask about the cylinders and not the entire algebra generated by them. Right? Note that I am very please by this answer since it says that absolute continuity need only be checked on a computable collection of subsets of $2^{\omega}$. – Quinn Culver Oct 04 '12 at 20:18
  • By the way, I'd now like to ask the same question(s) using the standard definition of $\ll$: is there a countable collection $\mathcal{C}$ of Borel sets such that $\mu \ll \nu$ iff $\nu C =0 \Longrightarrow \mu C=0$ for each $C \in \mathcal{C}$. Do you happen to know the answer to that? Should I ask a new question or just edit this one? (Note that the latter might require me to rescind my acceptance of your current answer.) – Quinn Culver Oct 04 '12 at 20:54
  • Isn't the cylinder already an algebra? For the other way, I think it also works, as a Borel set can be approximate innerly by a cylinder. So if a Borel set has $\nu$ measure $0$, each cylinder subset has measure $\mu$ equal to $0$. – Davide Giraudo Oct 06 '12 at 20:39
  • I think an algebra must be closed under (finite) unions, so no, the collection of cylinders is not an algebra (just take two incomparable strings). Not every Borel set can be approximated from within by a cylinder: for example the Lebesgue measure of the finite union of cylinders $[01] \cup [10]$ cannot be approximated by a single cylinder. – Quinn Culver Oct 06 '12 at 21:39
  • @QuinnCulver I understand your point. So what I did works for the collection of finite unions of cylinders, not for the cylinders themselves. I will think (much) more deeply on the problem. – Davide Giraudo Oct 06 '12 at 22:05
  • I am removing my acceptance of this answer in hopes that Question 1 gets some attention. – Quinn Culver Feb 05 '13 at 20:44
  • Would you please tell me if you think my answer to Question 1 is correct? – Quinn Culver Feb 08 '13 at 14:18
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Any measure that is atomless but not absolutely continuous with respect to the Lebesgue measure witnesses that the answer to Question 1 is negative.

If $\mu$ is atomless, it must satisfy the $\epsilon$-$\delta$ condition given for cylinders because otherwise, there is $\epsilon>0$ such that for all $n$ there is $\sigma$ of length $n$ (i.e. Lebesgue measure less than $2^{-n}$) such that $\mu([\sigma])> \epsilon$. But the collection of all such $\sigma$'s would then be an infinite finitely-branching tree and hence have a path (by König's lemma), which would be an atom for $\mu$.

For a concrete example, take the Bernoulli $(1/3, 2/3)$ measure $\mu$ (the measure that corresponds to the process of flipping a bias coin with probability 1/3 of heads). This measure is clearly atomless, but is not absolutely continuous with respect to Lebesgue measure since it says that the Lebesgue-null set of sequences without the same number of $1$'s as $0$'s has measure $1$.

Quinn Culver
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