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

Wisniewool 2022-07-12 Answered
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?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Mateo Carson
Answered 2022-07-13 Author has 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.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-06-14
if I know that a function f L 2 ( R d ) is supported in a compact ball in R d , how do I prove it is in L 1 ( R d )

From the square integrability I know that
for some constant C, then I know that f ( s u p p ( f ) ) is bounded, I need to show that it is closed as well to conclude that it is compact, but that will not be true in general if f is not continuous.

How am I supposed to start given what I have?
asked 2022-06-15
Let γ : [ 0 , 1 ] S 2 be a smooth curve. Consider the following set
t [ 0 , 1 ] s p a n ( γ ( t ) ) R 3
My question is: does this set have Hausdorff dimension at most 2, and as a result it has zero 3-dimensional Lebesgue measure?. Note that if the smoothness condition is removed then the answer is trivially false.
asked 2022-06-14
Let ( Σ , F , P ) be a probability space, let X be an F-random variable, let G be a sub-sigma algebra of F, let ω Ω, let A be the intersection of all the sets in G that contain ω, and suppose that A F. Then is it true that E ( X | G ) ( ω ) = E ( X | A )?
I just wrote this statement down based on my intuition of what conditional expectation means, and I want to verify if it’s actually true. Note that sigma algebras need not be closed under uncountable intersections, so A need not be an element of G. But I assumed A is at least an element of F so that E ( X | A ) is meaningful.
asked 2022-05-26
Let S , T , I be non-empty sets, for every i I let A i be a sigma-algebra on T, and let f : S T. Is it true that f 1 ( i I A i ) = i I f 1 ( A i )?

Remarks
1. It is known that f 1 ( i I B i ) = i I f 1 ( B i ) if B i T for every i I. But this result is not the same as what needs to be proved, since i I B i T, whereas i I A i P T.
2. I was able to show that f 1 ( i I A i ) i I f 1 ( A i ), so the question reduces to whether the converse containment is valid.
asked 2022-07-09
Let ( Ω , μ ) be a measure space. Suppose that ( f n ) converges almost everywhere to some function f where each f n belongs to L ( Ω ). Suppose that sup n f n < .
Do we have f L ( Ω ) ?
asked 2022-05-21
Need help:
Let ( Ω , F , P ) be a probability space. Suppose f : R R is some function such that f ( X ) is measurable for every real valued random variable X. I am curious if f ( X ) is necessarily σ ( X )-measurable. I tried to conclude with Doob-Dynkin lemma but to do we would need f to be B ( R ) measurable. Does someone has an idea or is this is false in general?
asked 2022-06-29
I wonder why uncertainties in angle measurement MUST be in radians. For example, I want to calculate the uncertainty in measuring the function y = sin ( θ ) when the angle is measured ± 1 degree. I do this using differential, so d y = cos ( θ ) d θ, now d θ = ± 1 degree is the error in θ. Now, all the course notes/ books I read says this must be converted in radians, even though the angle we use here is measured in degree. How come?