Measure as Metric

It turns out that a measure turns its domain into a sort of metric space, measuring the “distance” between two sets. So, let’s say $\mu$ is a measure on an algebra $\mathcal{A}$ , and see how this works.

First, we need to define the “symmetric difference” of two sets. This is the collection of points in one or the other set, but not in both:

$\displaystyle A\Delta B = (A\setminus B)\cup(B\setminus A)$

So now we define an extended real-valued function $d:\mathcal{A}\times\mathcal{A}\to\overline{\mathbb{R}}$ by

$\displaystyle d(A,B)=\mu(A\Delta B)$

I say that this has almost all the properties of a metric function. First of all, it’s clearly symmetric and nonnegative, so that’s two of the four right there. It also satisfies the triangle inequality. That is, for any three sets $A$ , $B$ , and $C$ in $\mathcal{A}$ , we have the inequality

$\displaystyle d(A,B)\leq d(A,C)+d(C,B)$

Indeed, points in $A\setminus B$ are either in $C$ or not. If not, then they’re in $A\setminus C$ , while if they are they’re in $C\setminus B$ . Similarly, points in $B\setminus A$ are either in $B\setminus C$ or $C\setminus A$ . That is, the symmetric difference $A\Delta B$ is contained in the union of the symmetric difference $A\setminus C$ and the symmetric difference $C\setminus B$ . And so monotonicity tells us that

$\displaystyle\mu(A\Delta B)\leq\mu((A\Delta C)\cup(C\Delta B))\leq\mu(A\Delta C)+\mu(C\Delta B)$

establishing the triangle inequality.

What’s missing is the assertion that $d(A,B)=0$ if and only if $A=B$ . But there may be plenty of sets with measure zero, and any one of them could arise as a symmetric difference; as written, our function $d$ is not a metric. But we can fix this by changing the domain.

Let’s define a relation: $A\sim B$ if and only if $\mu(A\Delta B)=0$ . This is clearly reflexive and symmetric, and the triangle inequality above shows that it’s transitive. Thus $\sim$ is an equivalence relation, and we can pass to the collection of equivalence classes. That is, we consider two sets $A$ and $B$ to be “the same” if $A\sim B$ .

This trick will handle the obstruction to $d$ being a metric, but only if we can show that $d$ gives a well-defined function on these equivalence classes. That is, if $A\sim A'$ and $B\sim B'$ , then $d(A,B)=d(A',B')$ . But $A\sim A'$ means $d(A,A')=0$ , and similarly for $B$ . Thus we find

$\displaystyle d(A,B)\leq d(A,A')+d(A',B')+d(B',B)=d(A',B')$

and, similarly

$\displaystyle d(A',B')\leq d(A',A)+d(A,B)+d(B,B')=d(A,B)$

and so the two are equal. We define the distance between two $\sim$ -equivalence classes by picking a representative of each one and calculating $d$ between them.

This relation $\sim$ turns out to be extremely useful. That is, as we go forward we will often find things simpler if we consider two sets to be “the same” if they differ by a set of measure zero, or by a subset of such a set. We will call subsets of sets of measure zero “negligible”, since we can neglect things that only happen on such a set.

March 24, 2010 - Posted by John Armstrong | Analysis, Measure Theory

10 Comments »

I happened to surf by your blog and saw your discussion about the measure symmetric difference metric. About 5 years ago I posted a mini-survey of this metric that you might be interested in.

http://tinyurl.com/yd3dhsg

Here’s an excerpt:

A related interesting result is given in Kato/Kanzo/Shinnosuke [11]. They consider the space that I’ll call N[0,1], which consists of *all* subsets of [0,1] modulo the same equivalence relation, where the distance is defined in the same way except that outer Lebesgue measure is used. They prove that M[0,1] is a perfect nowhere dense set in N[0,1]. In other words, the collection of measurable subsets of [0,1] forms a very tiny part of the collection of all the subsets of [0,1]!

[11] Nobuyuki Kato, Tadashi Kanzo, and Oharu Shinnosuke, “A note on the measure problem”, International Journal of Mathematical Education in Science and Technology 19 (1988), 315-318.

Comment by Dave L. Renfro | March 24, 2010 | Reply
Thanks, Dave. I’ll check it out (as should anyone else who’s curious).

I’m not sure how much I’ll use this metric space, though. I’m mostly talking about it here because I think it’s an interesting way of looking at things, and it’s a good motivation for the idea of negligible sets.

Comment by John Armstrong | March 24, 2010 | Reply
- It seems I overstated things. Although I do talk about this metric space, the main focus of that essay is on various “size results” about intersections of measurable sets. The measure symmetric difference metric enters the scene as an interesting way (due to Gillis, reference below) of proving an intersection result by applying the Cantor-Bendixson theorem to this measure symmetric difference metric space.
  
  Joseph E. Gillis, “Note on a property of measurable sets”, Journal of the London Mathematical Society 11 (1936), 139-141.
  
  Comment by Dave L. Renfro | March 24, 2010 | Reply
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please,i want answer to question says the symmetric difference of two measurable sets is measurable

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