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$F$ is a finite measure on $(X,A)$

$a$ and $b$ belong to $A$

show that $F(a \cup b)=F(a)+F(b)-F(a \cap b)$

I have no idea how to approach this question. Any assistance would be appreciated.

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Venn diagrams? Honestly, I don't know enough about measure theory to give a proper answer, but that looks like the definition of union in the context of measure, so I imagine one could approach it as one approaches normal unions. Then again, I could be way off base. –  Jack Henahan Jun 19 '11 at 7:33
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2 Answers

Hint: write $a \cup b$ as the following disjoint union: $$ a \cup b = [a - (a \cap b)] \cup (a \cap b) \cup [b - (a \cap b)]. $$

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The case where $F(X)=0$ is trivial; the case where $F(X) > 0$ is essentially nothing more than ${\rm P}(a \cup b) = {\rm P}(a) + {\rm P}(b) - {\rm P}(a \cap b)$ from probability theory. –  Shai Covo Jun 19 '11 at 10:57
    
It is no easier to prove in "probability theory" than in "measure theory". –  GEdgar Jun 19 '11 at 13:29
    
@GEdgar, of course, it is essentially the same (but the comment might be useful for those who don't know measure theory). –  Shai Covo Jun 19 '11 at 13:44
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Note that $ a = (a\cap b^c) \cup (a \cap b)$
so $$F(a)=F(a \cap b^c) + F(a \cap b)$$
Also, $a \cup b = (a\cap b^c) \cup b$, so $F(a \cup b)= F(a\cap b^c) + F(b)$. Hence $$ F(a \cup b) = ...$$

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