- #1
Oxymoron
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I want to prove that if E is a subset of F and both E and F are measurable, then m(E) </= m(F). (where </= is less than or equal to).
Now I figured that I'd use one of the axioms for a measure to prove this, namely
If A_i are measureable and disjoint A_n n A_m = {} for n not equal to m, then
m(U A_i) = sum m(A_i)
(for some reason tex is not working for me anymore.)
But that requires me to form the following collection of sets:
E*_1 = E
E*_2 = F - E
E*_3 = E*_4 = ... = emptyset
to ensure that each E*_i are measureable and disjoint. Unfortunately I have no justification for doing so, only that I've seen it done before. In fact, I am not even sure of how to continue. Any help would be much appreciated.
Now I figured that I'd use one of the axioms for a measure to prove this, namely
If A_i are measureable and disjoint A_n n A_m = {} for n not equal to m, then
m(U A_i) = sum m(A_i)
(for some reason tex is not working for me anymore.)
But that requires me to form the following collection of sets:
E*_1 = E
E*_2 = F - E
E*_3 = E*_4 = ... = emptyset
to ensure that each E*_i are measureable and disjoint. Unfortunately I have no justification for doing so, only that I've seen it done before. In fact, I am not even sure of how to continue. Any help would be much appreciated.