# Symmetric Difference as a Metric

I'm currently working on a homework question, and I am stuck. I copied this off the whiteboard, so it is very possible that I made a mistake in transcribing it. Of course I may be misinterpreting something, and if that is the case, I would like some clarification.

We are given that $(X,M,\mu)$ is a measure space. For $A,B\subset X$, we have that $A\triangle B = (A\setminus B)\cup (B\setminus A)$. We then have that $d(C_1,C_2) = \mu(A\triangle B)$ for $A\subset C_1$ and $B\subset C_2$. We are supposed to prove that $d$ is a well-defined distance function, but I don't see how that is possible. I feel like I could complete the proof if $d(A,B)=\mu(A\triangle B)$, but that is not what I am given. Any hints would be appreciated.

Edit: Yes, $C_1, C_2\in M$.

Edit: Haha, the last guess was right. It's 4:00AM where I'm living currently, but I found someone who was awake and he gave me the correct answer. Thanks for your help everyone.

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We may need some $\sup$ at the def. of $d$, and, are these $C_1,C_2\in M$? – Berci Oct 4 '12 at 23:16
I think that @Berci is right: the definition should be $$d(C_1,C_2)=\sup\{\mu(A\triangle B):A\subseteq C_1\text{ and }B\subseteq C_2\}\;.$$ – Brian M. Scott Oct 4 '12 at 23:20
@Brian: But that would make $d(C,C)=\mu(C)$ for any measurable $C$ (take $A=C$ and $B=\varnothing$) -- which I don't think is allowed for a metric. – Henning Makholm Oct 5 '12 at 0:13
If $C_1,C_2$ are measurable, then the definition should be just $d(C_1,C_2)=\mu (C_1\triangle C_2)$ (at least that's the one I know...). Not sure what $A,B$ are for here. – tomasz Oct 5 '12 at 0:34
After some thought, maybe it's supposed tobe $A\in C_1$ and $B\in C_2$, with $C_1, C_2$ equivalence classes of the relation $A\sim B \iff \mu(A\triangle B)=0$ (so that it's actually a metric, and not a premetric). – tomasz Oct 5 '12 at 0:36

If $d$ should be a bona fide metric, one has to assume that $\mu$ is finite, and identify the sets that agree up to a null set. After that, the first two axioms are clear, and only the triangle inequality remains. One can prove the triangle inequality by playing with Venn diagrams, but there's also another approach: the map $$A\mapsto \chi_A$$ embeds the space of sets (rather, their equivalence classes) into $L^1(X,M,\mu)$. The latter has a metric coming from its norm, and it should be clear that $d$ agrees with this metric.
But here $A$ is arbitrary subset of $X$, may not be measurable set. – Ram Nov 10 '13 at 18:32