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Let $(X, \Sigma, \mu)$ be a measure space.

We can define a pseudometric $d$ on $\Sigma$ in the following way:

$$d(A, B) = \mu(A\bigtriangleup{}B)$$

where $A\bigtriangleup{}B = (A\cup{}B)\setminus{}(A\cap{}B)$ is the symmetric difference.

Does this metric have a name? What does the topology induced by $d$ on $\Sigma$ tell us about the measure?

Rasmus
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pokraka
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    It's just the standard (pseudo)metric induced by the measure; normally this is defined on the quotient measure algebra (modding out the null sets), making it a metric on that. It's used to define a separable measure: iff this metric space is separable. – Henno Brandsma Jun 08 '11 at 19:34

3 Answers3

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Assuming $\mu(X)<\infty$, this is the restriction of the $L^1$ metric on $L^1(X)$ to the subspace consisting of characteristic (a.k.a. indicator) functions. Clearly this is a closed (topological) subspace, and therefore $\Sigma$ is a complete metric space under this metric.

As far as I know, this metric itself doesn't have a name. In some sense, it should give precisely the same information about the measure that you can get from the space of $L^1$ functions.

Edit: As Theo pointed out, the last sentence isn't quite correct, since the subspace of characteristic functions cannot actually be recognized from within $L^1$. For example, $L^1([0,1])$ and $L^1([0,2])$ are isomorphically isometric via the map $f(x) \mapsto f(x/2)/2$, but the corresponding $\Sigma$'s are not isometric. In general, the total measure $\mu(X)$ of the underlying space cannot necessarily be recovered from $L^1(X)$, but $\mu(X)$ is the diameter of $\Sigma$ as a metric space.

Edit #2: Some further comments:

  1. Note that $\Sigma$ is a topological group under $\bigtriangleup$, and is therefore homogeneous. As a result, we can assume that we know which point of $\Sigma$ represents the empty set $\emptyset$.

  2. Let $\sim$ be the equivalence relation $$ A\sim B \qquad\Leftrightarrow\qquad \mu(A \bigtriangleup B) = 0. $$ Since $\Sigma$ is homogeneous, every equivalence class under $\sim$ has the same cardinality as the equivalence class of $\emptyset$. Thus, we can use the pseudometric structure on $\Sigma$ to determine the cardinality of the collection of measure-zero sets.

  3. Assuming $\mu(X) < \infty$, the metric structure on $\Sigma$ can also be used to reconstruct the $\sigma$-algebra structure on $\Sigma/\sim$. In particular, if $A,B\in \Sigma/\sim$, we can tell whether $A\subset B$ (modulo sets of measure zero) by checking whether $$ d(\emptyset,A) + d(A,B) = d(\emptyset,B)\text{,} $$ and this lets us reconstruct the notions of countable union and complement.

  4. According to point #3, the metric structure on $\Sigma$ lets us reconstruct $\Sigma/\sim$ as a measure algebra (see this reference for the definitions used in this paragraph). Moreover, it should be clear that $\Sigma/\sim$ and $\Sigma'/\sim$ are conjugate as measure algebras if and only if they are isometric as metric spaces.

We conclude that, in the case where $\mu(X) < \infty$, the metric structure on $\Sigma$ contains only the following information:

(1) The cardinality of the collection of sets of measure zero and

(2) The conjugacy class of $(X,\Sigma,\mu)$ as a measure algebra.

According to this reference, any two separable, non-atomic probability spaces are measure algebra conjugate. Therefore, as long as $(X,\Sigma,\mu)$ is separable and non-atomic, the only information contained in point (2) is the total measure $\mu(X)$ of $X$.

Jim Belk
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    Jim: This space has much more information than the metric space $L^1$ alone. Note that $d(\phi,A) = \mu(A)$, so the measure is stored in there. However, for two mutually absolutely continuous measures $\mu$ and $\nu$ the spaces $L^1(\mu)$ and $L^1(\nu)$ are isometrically isomorphic via multiplication by the Radon Nikodym derivatives. Thus your last statement must involve more information on $L^1$ than purely metric information. The metric space $(\Sigma,d)$ has much more structure (it is a Boolean algebra with a topology) that can be exploited differently than $L^1$ as a Banach space alone. – t.b. Jun 09 '11 at 00:21
  • @Theo: I see, you can't actually recognize the characteristic functions within $L^1$ using the metric, so there's more information contained in $\Sigma$ than there is in $L^1$. I'll edit my post to reflect this. – Jim Belk Jun 09 '11 at 00:36
  • Your addition is very nice and shows that in the separable case my point is essentially moot. – t.b. Jun 09 '11 at 02:54
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The function $d$ is sometimes called the Fréchet-Nikodym metric; for example, see section 1.12(iii) on page 53 of Measure Theory: Volume 1 by V. Bogachev.

Added: Google Books led me to the Encyclopedia of Distances by Michel M. Deza, Elena Deza. There the metric is called the symmetric difference metric, or the Fréchet-Nikodym-Aronszyan metric, or the measure metric.

  • Does Bogachev give references to the works of Fréchet and Nikodym? (Google doesn't let me look) – t.b. Jun 08 '11 at 23:47
  • @Theo Bogachev lists 16 works of M. Frechet and 10 works of O. Nikodym in the back of his book. However, section 1.12(iii) Metric Boolean Algebra doesn't have any references in it. So it's not clear exactly which of these works discuss the metric, or how it got its name. –  Jun 09 '11 at 01:04
  • Thanks! That's what I feared... So I'll have to dig in the library if I really want to know. I'll post a comment here if I should happen to find something worth mentioning. – t.b. Jun 09 '11 at 01:10
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Separability of this spaces gives separability of the $L^p$ spaces, $1\le p < \infty$.

ncmathsadist
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