# Measure theory and the symmetric difference

1. Aug 5, 2008

### nyq_guru

Hi,

I'm currently trying to teach myself some measure theory and I'm stuck on trying to show the following:

Let $$(X,M,\mu)$$ be a finite positive measure space such that $$\mu({x})>0$$ $$\forall x \in X$$ . Set $$d(A,B) = \mu(A \Delta B)$$, $$A,B \in X$$. Prove that $$d(A,B) \leq d(A,C) + d(C,B)$$ .

Could someone perhaps help me on the way? I've tried various approaches (mostly "brute force" with different re-writings of both sides of the inequality) but every time there is some term left that doesn't quite fit. Feels like it's actually quite easy and that I'm just missing the point.

Thanks in advance, oh and btw if this does not come out in the correct tex format, how do I achieve that? My first time here

Last edited: Aug 5, 2008
2. Aug 6, 2008

### morphism

Have you tried proving that $A \Delta B \subset A \Delta C \cup C \Delta B$?

3. Aug 6, 2008

### nyq_guru

Actually no, but I think I see where you're getting at. If I can prove this, it would imply that $$\mu(A \Delta B) \leq \mu((A \Delta C) \cup (C \Delta B))$$.
Now I tried to prove the relation you mentioned, but I can't seem to do it. The expression for the $$A \Delta C \cup C \Delta B$$ is to nasty for me this early in the morning, especially since I'm still not comfortable enough with set algebra. However, with Venn diagrams it is easy to see.
Now, if this relation is proved, I simply use that (since it is a finite positive space) $$\mu((A \Delta C) \cup (C \Delta B)) = \mu(A \Delta C) + \mu(C \Delta B) - \mu((A \Delta C) \cap (C \Delta B))$$. Since the last term is >0, it is clear that the inequality holds, right?

4. Aug 6, 2008

### HallsofIvy

Haven't had your coffee yet? I'm not sure what you mean by "set algebra" but I would just go back to the basic definitions: to prove
$A \Delta B \subset A \Delta C \cup C \Delta B$

If x is in $A \Delta B$ then x is in A but not in B. Now look at two cases:

case 1: x is in C. Then x is in $C \Delta B$

case 2: x is NOT in C. Then x is in $A \Delta C$

5. Aug 6, 2008

### nyq_guru

Now forgive me if I'm wrong, but doesn't $$x \in A \Delta B$$ imply that x is in either A or B, but $$x \notin A \cap B$$?
From drawing a Venn diagram for $$A \Delta C \cup C \Delta B$$ I believe that this relation can be written $$(A \cup B \cup C)$$ \ $$(A \cap B \cap C)$$. If this is correct, obviously any $$x \in A \Delta B$$ will also be in $$A \Delta C \cup C \Delta B$$. However, if this is the case I would like to show that $$A \Delta C \cup C \Delta B = (A \cup B \cup C)$$ \ $$(A \cap B \cap C)$$, not just draw a diagram.
Using the basic definitions would yield that $$x \in A \Delta B$$ if $$x \in A$$ or $$x \in B$$ but $$x \notin A \cap B$$. Similarly, $$x \in A \Delta C \cup C \Delta B$$ implies that (if the equality earlier is true) that $$x \in A$$, $$x \in B$$ or $$x \in C$$ but $$x \notin A \cap B \cap C$$, which I believe shows that $$A \Delta B \subset A \Delta C \cup C \Delta B$$

6. Aug 6, 2008

### morphism

HallsofIvy's proof is almost complete. He showed that $A \backslash B \subset C \Delta B \cup A \Delta C$. Symmetrically, we have that $B \backslash A \subset C \Delta A \cup B \Delta C$, and the result now follows.

7. Aug 6, 2008

### nyq_guru

Something went seriously wrong with my last post, everything just became a blur. The expressions aren't even in the correct places. Oh well, I'll fix it when I get home. morphism, thanks for pointing out that way to complete the proof. I still can't let go of the thought of rewriting the expression for $$A \Delta C \cup C \Delta B$$ though :-).

8. Aug 6, 2008

### HallsofIvy

You are right. I completely missed that: I was thinking "difference" rather than "symmetric difference". And I had had my coffee!