E c s it suffices to prove the measurability of sets

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E c S , it suffices to prove the measurability of sets E such that E ( -∞ , 0]. Assume thus E ( -∞ , 0]. For every subset A of R , we’ll have sup( A E ) 0, thus μ * ( A E ) = 0. Assume first that A contains some x , x > 0. Then it is a simple exercise to prove that sup A = sup( A (0 , )) > 0, thus μ * ( A ) = sup( A (0 , )). We also have A (0 , ) = A E c (0 , ) and 0 < sup( A (0 , )) = sup( A E c (0 , )) = sup( A E c ) = μ * ( A E c ); thus μ * ( A ) = μ * ( A E c ) if a (0 , ) 6 = . If A (0 , ) = , then A ( -∞ , 0] and μ * ( A ) = 0, μ * ( A E c ) = 0 μ * ( A ) = 0 , so in all cases we have μ * ( A ) = μ * ( A E c ) Thus μ * ( A ) = 0 + μ * ( A ) = μ * ( A E ) + μ * ( A E c ) proving E is measurable. To conclude, we have to see that if E / S , then E is not measurable. Assume E / S . Then E 6⊂ ( -∞ , 0], thus there is x E , x > 0. But E 6 = F (0 , ) for a set F ( -∞ , 0]; i.e., E 6⊃ (0 , ), hence there is y > 0, y / E . Let A = { x, y } . Then A E = { x } , hence μ * ( A E ) = sup( A E ) = x , μ * ( A E c ) = sup( A E ) = y , and μ * ( A E ) + μ * ( A E c ) = x + y > max( x, y ) = μ * ( A ). Thus E / S . (c) μ * ( A ) = 1 if A is uncountable, 0 if A is countable (or finite). Solution. The empty set is finite thus μ * ( ) = 0; OM1 holds. If A B , if B is uncountable, then μ * ( A ) 1 = μ * ( B ); if B is countable, then so is A and μ * ( A ) = μ * ( B ) = 0. We see OM2 holds. If A n R for all n N , if A n is uncountable for some n N , then n =1 μ * ( A n ) μ * ( A n ) = 1 μ * ( S n =1 A n ). On the other hand, if all A n ’s are countable, so is their union (a countable union of countable sets is countable) and n =1 μ * ( A n ) μ * ( A n ) = 0 = μ * ( S n =1 A n ). Let S consist of all sets that are either countable or that have countable complement. We claim this is precisely the σ -algebra of measurable sets. Let E S . Suppose A R . Assume first A is countable. Then both A E and A E c are countable and μ * ( A ) = 0 = μ * ( A E ) + μ * ( A E c ) . On the other hand if A is uncountable then precisely one of A E, A E c is uncountable, the other one countable. In fact, one has to be uncountable, because A is uncountable. But one has to be countable, because one of E, E c is countable. Thus μ * ( A ) = 1 = 1 + 0 = μ * ( A E ) + μ * ( A E c ) . This proves that all sets of S are measurable. Assume now that E / S . Since R is uncountable, the only way this is possible is if both E and its complement are uncountable. Then 1 = μ * ( R ) < 1 + 1 = μ * ( R E ) + μ * ( R E c ) , showing that E is not measurable. 2. Prove: If E, F are measurable, if E F and if m ( E ) < , then m ( F \ E ) = m ( F ) - m ( E ). Why do we need to assume m ( E ) < ? Proof. If E, F are measurable, so is E \ F and (in case E F ) F = E ( F \ E ). Since this is a disjoint union, m ( F ) = m ( E ) + m ( F \ E ). If m ( E ) < we can subtract m ( E ) to get m ( F \ E ) = m ( F ) - m ( E ). If m ( E ) = , so is m ( F ) and m ( F ) - m ( E ) is undefined. The set F \ E could be a set having finite or infinite measure.
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