Let X be a compact Hausdorff space and &#x03BC;<!-- μ --> be a complex Borel measure on X

Wisniewool

Wisniewool

Answered question

2022-07-12

Let X be a compact Hausdorff space and μ be a complex Borel measure on X such that X f   d μ 0 for every f C ( X ) with f 0. Then show that μ is non-negative.
Could anyone give some suggestion in this regard?

Answer & Explanation

Mateo Carson

Mateo Carson

Beginner2022-07-13Added 15 answers

Let E X Borel. Let ϵ > 0. By regularity, we can choose a compact subset K X and an open subset U X with K E U and | μ | ( U K ) < ϵ. By Urysohn's lemma, we can pick f C ( X , [ 0 , 1 ] ) with χ K f χ U . Then
X | χ E f | d | μ | | μ | ( U K ) < ϵ .
This shows that the function χ E can be approximated by positive continuous functions in the L 1 ( | μ | )-norm.
Next, let E X be a Borel subset. Choose a sequence of positive continuous functions { f n } n = 1 C ( X , [ 0 , [ ) such that
X | f n χ E | d | μ | n 0.
Hence,
| X f n d μ X χ E d μ | X | f n χ E | d | μ | n 0
and we deduce that
μ ( E ) = X χ E d μ = lim n X f n d μ 0 0.
Since E was an arbitrary Borel set, we conclude that μ is a positive (finite) measure.

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