For both parts b c we might try to understand first

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For both parts (b), (c) we might try to understand first what needs to hold for Borel sets to be measurable, in particular how does the proof in Royden for the Lebesgue outer measure work. Some of the statements below should be clear (in fact, they should all be clear, but I don’t quite believe in miracles). (a) It suffices to prove that all sets of the form ( a, ) are measurable, for all a R . (b) To prove that a set of the form ( a, ) is measurable it is necessary and sufficient to show that for all subsets A of R , setting A 1 = A ( a, ), A 2 = ( -∞ , a ], then (1) μ * ( A ) μ * ( A 1 ) + μ * ( A 2 ) .
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(c) Given the definition of μ * ( A ) as an infimum, proving (1) is equivalent to proving: If { I n } is an arbitrary countable collection of open intervals covering A , then (2) μ * ( A 1 ) + μ * ( A 2 ) X n =1 ˜ ( I n ) . We can keep following Royden for a while longer. To prove (2) Royden introduces (and so do we) I 1 n = I n ( a, ), I 2 n = I n ( -∞ , a ]; then A 1 S n =1 I 1 n , A 2 S n =1 I 2 n . Because we have an outer measure, we have μ * ( A 1 ) X n =1 μ * ( I 1 n ) , μ * ( A 2 ) X n =1 μ * ( I 2 n ) thus (3) μ * ( A 1 ) + μ * ( A 2 ) X n =1 ( μ * ( I 1 n ) + μ * ( I 2 n ) ) . Royden completes the proof of (2) by using a result he proved before, namely that m * ( I ) = ( I ) for all intervals I , plus the obvious fact that ( I 1 n )+ ( I 2 n ) = ( I n ). If we had the similar results for μ * , ˜ instead of m * , ‘ , respectively, we could wrap up things just as Royden. It turns out that μ * ( I ) = ˜ ( I ) for all intervals I is true if f is continuous (which is why I thought continuity was essential) but fails to be true if f has jumps. Noticing that the intervals I n , I 1 n are open, while I 2 n is of the form ( · , · ], we only will have to figure out the relation between μ * and ˜ for intervals of this type. So we leave things on hold at (3) and take a detour to explore these relations. The ideas used in doing all this are all already in Royden; I am merely modifying them to apply in the case we are considering. Claim 1 Let I be an interval of endpoints a, b , a < b . Then (4) f ( b - ) - f ( a +) μ * ( I ) f ( b +) - f ( a - ) . (If a = -∞ , by definition f ( -∞- ) = f ( -∞ ); if b = , f ( ) = f ( +).) To see that μ * ( I ) f ( b +) - f ( a - ) we notice that one can find a sequence { J n } of open intervals, J n = ( c n , d n ) where c n a < b d n and { c n } goes to a , d n to b as n → ∞ . In fact, if a / I , so I is open on the side of a , just set c n = a for all n . If a I , then a R ( a is not -∞ ); set c n = a - 1 /n . Similarly, if b / I , set d n = b for all n ; otherwise set d n = b + 1 /n . Now I J n (which can be interpreted as I J n ∪ ∅ ∪ ∅ ∪ · · · ) so that by the definition of μ * we see that μ * ( I ) ˜ ( J n ) = f ( d n ) - f ( c n ) for all n N . Letting n → ∞ we get μ * ( I ) f ( b +) - f ( a - ) , the second inequality in (4). To prove the first inequality in (4), we may assume that μ * ( I ) < , otherwise the inequality is obvious. Given then ² > 0, there is a countable collection { J n } of open intervals such that I [ n =1 J n and such that if J n = ( c n , d n ) for n N , (5) X n =1 ˜ ( J n ) = X n =1 ( f ( d n ) - f ( c n )) μ * ( I ) + ².
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