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Dwllane4

Dwllane4

Answered question

2022-06-13

Let E be a metric space and M ( E ) denote the set of finite signed measures on ( E , B ( E ) ). Remember that C C b ( E ) is called separating for F M ( E ) if
(1) μ , ν F : ( f C : f d ( μ ν ) = 0 ) μ = ν .
Let M + ( E ) and M 1 ( E ) denote the set of finite and probability measures on ( E , E ), respectively.

If C is separating for M 1 ( E ), does it already follow that C is separating for M ( E )?

I'm not sure whether I'm missing something or not, but I think this should be true, since every finite signed measures can be written as a difference of two finite measures and every nontrivial finite measure can be normalized to a probability measure.

Answer & Explanation

kejohananws

kejohananws

Beginner2022-06-14Added 19 answers

True if the constant function 1 belongs to C . In this case f d ( μ ν ) = 0 for all f C implies μ = ν even μ , ν are just positive finite measures (not necessarily probability measures). [This is because μ ( E ) = ν ( E )].
Suppose f d ( μ ν ) = 0 for all f C . Write μ = μ 1 μ 2 , ν = ν 1 ν 2 where μ 1 , μ 2 , ν 1 , ν 2 are finite positive measures. Then f d ( ( μ 1 + ν 2 ) ( μ 2 + ν 1 ) ) = 0. It follows that μ 1 + ν 2 = μ 2 + ν 1 from which we get μ = ν.

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