Simple (I think?) measure theory question

[AxiomOfChoice]

If you have two measurable sets $A$ and $B$ (not necessarily disjoint), is there an easy formula for the measure of the difference, $m(A-B)$?

 

[g_edgar]

$$m(A-B) = m(A) - m(A\cap B)$$
or, slightly better since it holds even if $$m(A) = \infty$$,
$$m(A-B) + m(A\cap B) = m(A)$$

 

[jambaugh]

By set difference do you mean symmetric difference?
$$A-B= A\cap \overline{B} \cup \overline{A}\cap B$$
where overline is set complement?

g_edgar's formulas are for:
$$A-B = A \cap \overline{B}= \{ x | x\in A \, \&\, x\not\in B\}$$

I'll use ~ for symmetric difference and then:
$$m(A \sim B) = m(A) + m(B) - m(A\cap B)$$
Its just a matter of looking at a Venn diagram and thinking of areas as the measure.

 

[AxiomOfChoice]

$$m(A-B) = m(A) - m(A\cap B)$$
or, slightly better since it holds even if $$m(A) = \infty$$,
$$m(A-B) + m(A\cap B) = m(A)$$

Thanks! But can you explain why this is this justified? There is a corollary in my textbook that gives $m(B-A) = m(B) - m(A)$ if $A\subseteq B$. Do we have $S-T = S - (S\cap T)$ for any sets $S$ and $T$? If we do, I'm satisfied...

 

[Office_Shredder]

Do we have $S-T = S - (S\cap T)$ for any sets $S$ and $T$? If we do, I'm satisfied...

Of course you do! S-T is the points in S that are not in T. But the only points that can be removed from S are the points in T that are also in S. And $S-(S \cap T)$ is precisely the points in S except for the points in T also in S. It's nearly a semantic proof

 

[jambaugh]

Informally you see it in a Venn diagram.
Formally you look at the additivity property of measures, i.e. the measure of the union of disjoint sets is the sum of the measures of the pieces.

When dealing with any combination of unions and/or intersections simply break the set in question down into its smallest** pieces.

Example: Given initial sets A, B, and C you can break A up into A = ABC U ABC' U AB'C U AB'C' (where I'm using AB for A intersect B and B' for complement of B.)

**(smallest in terms of not further divisible by intersection with a given set or set's complement.)

This is very very straightforward stuff. Again draw a Venn diagram! I think you may be over thinking this.