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Define the following pseudo-metric on the set of measurable subsets of $R$:

$$D(A,B)=\operatorname{Length}((A\setminus B) \cup (B\setminus A)),$$

i.e., the distance between $A$ and $B$ is the measure of their symmetric distance.

Is the set of measurable subsets of $R$ closed under this metric?

After reading a recent answer by Conifold, Compactness of a set of partitions, showing that the subspace of measurable functions is not closed under another the Tychonoff topology, my guess was that the set of measurable subsets is also not closed. So I tried to look for a counter-example consisting of:

  • A sequence of sets $A_1,A_2,...$, all measurable;
  • A non-measurable set $L$;

Such that the sequence $A_i$ converges to $L$ under $D$. But I am not exactly sure how to define this convergence. Initially I wrote: $M(A_i\setminus L \cup L\setminus A_i)$ converges to 0, but following a comment by GEdgar, I am not sure I can write this because $M$ is not defined on $L$.

Here is my attempt:

Define a sequence of finite sets that converges the set of rational numbers:

  • Define some linear ordering on the rational numbers in $[0,1]$.
  • For every $i>0$, define $Q_i$ as the rational numbers up to and including the $i$th rational number in that ordering.

Now define the sets $A_i$ using the "Vitali set" construction but taking $Q_i$ instead of $Q$.

Every $Q_i$ is finite, so the $A_i$'s are measurable. However, $Q_i$ converges to $Q$ so $L$ is a Vitaly set which is not measurable.

I am not convinced by this counter-example... can you provide a better one?

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Erel Segal-Halevi
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  • So, you have $M$ defined for non-measurable sets? I assume you do, since you wrote $M(A_i\setminus L \cup L\setminus A_i)$. – GEdgar Jun 23 '14 at 13:21
  • I think this might be related – DanZimm Jun 23 '14 at 13:24
  • I have to admit that I'm always finding it strange and amusing when someone gives a formatted link to Wikipedia and has a typo in the formatting, which is really just repeating the title. To be more to the point, Vitali and not Vitaly. :-) – Asaf Karagila Jun 30 '14 at 19:40

1 Answers1

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Define the distance between any two real sets $A,B\subset\Bbb R$ as $d(A,B)=\lambda^*(A\Delta B)$, where $\lambda^*$ is the Lebesgue outer measure and $\Delta$ is the Symmetric difference.

This turns the power set of $\Bbb R$ modulo null sets into a metric space.

The question you are asking is whether the $\sigma$-algebra of Lebesgue measurable sets is a closed subset of this metric space.

The answer is affirmative.

Suppose $A_n\rightarrow A$, i.e., $$\lambda^*(A\Delta A_n)\rightarrow 0$$ where $A_n$ is measurable, i.e., for any $X\subset\Bbb R$, $$\lambda^*(X)=\lambda^*(X\cap A_n)+\lambda^*(X\cap A_n^c)$$ (where $Z^c=\Bbb R \backslash Z$ is the complement of set $Z\subset \Bbb R$).

The proof consists of the following exercises:

  1. $\lambda^*(A^c\Delta A_n^c)\rightarrow 0$
  2. $\lambda^*(X)\le\lambda^*(X\cap Y)+\lambda^*(X\cap Y^c)$ for all $X,Y\subset\Bbb R$
  3. $\limsup \lambda^*(X_n)\ge \lambda^*(X) $ if $X_n\rightarrow X$
sds
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