2
$\begingroup$

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.

  • 0
    We may need some $\sup$ at the def. of $d$, and, are these $C_1,C_2\in M$?2012-10-04
  • 1
    I think that @Berci is right: the definition should be $$d(C_1,C_2)=\sup\{\mu(A\triangle B):A\subseteq C_1\text{ and }B\subseteq C_2\}\;.$$2012-10-04
  • 0
    @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.2012-10-05
  • 0
    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.2012-10-05
  • 2
    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).2012-10-05
  • 0
    Steve, if you now have the correct answer, why not post it as an answer (and then, sometime later, accept it)?2012-10-05

1 Answers 1