# The Unapologetic Mathematician

## 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 | Analysis, Measure Theory | 10 Comments