I read in an old book the following example of a measure: For the set M = <mrow class="MJX

lilmoore11p8 2022-07-14 Answered
I read in an old book the following example of a measure: For the set M = Q [ 0 , 1 ] denote with S the set system of subsets of M of the form Q I, where I is any interval in [0,1]. Let us define the function μ : S R as follows: for any set A S of the form A = Q I we set μ ( A ) = ( I ) = b a ..
Then it said without proof that μ is finitely additive, but not σ-additive.
As I did not get why I tried to prove it by myself and I tried to show that S is a semi-ring, I guess that is important before I start with the other proof.
We have Q and furthermore ( Q I 1 ) ( Q I 2 ) = Q I 1 I 2 and the union of two closed intervals is either an interval or the disjoint union of two intervals. Then ( Q I 1 ) ( Q I 2 ) is also an interval or the disjoint union of two intervals.
Now the proof. I do not quite understand how it cannot be σ-additive. Does it have something in common with Cantor sets? I don´t know how to start the proof here. Any help or explanation (maybe an idea for the beginning of a proof) is appreciated. If there is a proof...
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Answers (1)

Danika Rojas
Answered 2022-07-15 Author has 9 answers
If μ is σ-additive then for disjoint suitable sets A n it must satisfy:
μ ( n = 1 A n ) = n = 1 μ ( A n )
From this it can be deduced that for suitable sets B n (not necessarily disjoint) it must satisfy:
μ ( n = 1 B n ) n = 1 μ ( B n )
Now for every r M choose an interval I r with r S r = Q I r S .
If μ is indeed σ-additive then:
1 = μ ( M ) = μ ( r M S r ) r M μ ( S r ) = r M l ( I r )

However it is easy to choose the intervals I r in such a way that:
r M l ( I r ) < 1
and doing so we run into a contradiction.

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