I was wondering: a measure &#x03BC;<!-- μ --> is a function that takes a set of numbers S

pouzdrotf 2022-07-14 Answered
I was wondering: a measure μ is a function that takes a set of numbers S R n and assign a non-negative number to it. I'm summary: μ : S R + Does any of you know if there are measures like this:
μ : S R + 2
The final aim is being able to distinguish between countable dense sets and non dense sets, so a subset S Q can have a measure greater than zero if it's dense, at least in one of the dimensions of μ.
Many thanks in advance!
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)

Tucker House
Answered 2022-07-15 Author has 9 answers
One can certainly consider vector valued measures. Just take two measures μ 1 , μ 2 on the same measure space and set μ ( E ) = ( μ 1 ( E ) , μ 2 ( E ) ) R + 2 .

We have step-by-step solutions for your answer!

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 2022-06-25
I would like to know, why:
If X is a subspace of L p ( G ) such that X ¯ L p ( G ), then there exists g L q ( G ), 1 p + 1 q = 1 such that
G f ( x ) g ( x ) d x = 0 ; f X ¯ .
Where G is a locally compact group with Haar measure d x.
asked 2022-05-27
I have to prove the following:
Let X , Y be countable sets. Show that P ( X ) P ( Y ) = P ( X × Y ).
I'm not sure if in my case, P ( X ) P ( Y ) is a product σ-algebra and thus defined as P ( X ) P ( Y ) = σ ( { B 1 × B 2 | B 1 P ( X ) , B 2 P ( Y ) } ) .
I know that the power set of a set is a σ-algebra. So, it would make sense that P ( X ) P ( Y ) is a product σ-algebra as defined above.
Can somebody confirm this or explain it if it's not the case?
asked 2022-07-13
Let B([0,1]) be the borel set on [0,1] and suppose that μ is a finite measure on B([0,1]). Then, define A = [ 0 , 1 ] { x ; μ ( { x } ) > 0 } . I have shown that if A is countable, then [0,1]∖A is a separable metric space (because it is a subset of a separable metric space). Now if ( x n ) n is a dense sequence in [0,1]∖A, then I want to show that it is also dense in [0,1]. How can I prove it?
asked 2022-07-26
Convert 315 degree to radian measure in terms of pi.
asked 2022-07-01
Let I 1 , I 2 . . . be an arrangement of the intervals [ i 1 2 n , i 2 n ) in a sequence. If X n = n on I n and 0 elsewhere then s u p n E X n < but P ( sup n X n < ) = 0. My basic space is [0,1] with Lebesgue measure.
A := { sup n X n < } = { ω Ω : M > 0 , n N , X n ( ω ) M }
hence the complement is
A ¯ = { ω Ω : M N , n N , X n ( ω ) > M } = M N n N { X n > M } = M N n M { X n > M }
How do i go on from here?
asked 2022-06-14
Assume X is a continuous random variable which have a density function. Assume Y _ is a random vector which also have a density function. And finally assume we have a joint density function of the random variable and the random vector.
How can I find a function ϕ, such that E [ | X ϕ ( Y _ ) | | Y _ ] would be minimal?
I'm not sure where to start. I do know that the asnwer should be that ϕ ( Y _ ) should be a median of F X | Y _ , but Im not sure how to prove it. Any help would be appreciated.
Thanks in advance.
asked 2022-06-21
Let us say that we have some set of independent random variables: X = { X i } i = 1 n defined over a probability space of   ( Ω , F , P ). I want to understand whether the following holds:
If F X = σ ( { X 1 , X 2 , , X n } ) , then is it, in general, true that: | F X | = i I n | σ ( { X i } ) | ?
As we know, as X consists of independent R.V.-s, then we can state that i , j I n : σ ( { X i } ) and σ ( { X j } ) are independent. But how to proceed from this fact to the split of the cardinality of F X ?
I would appreciate any help, thank you in advance!

New questions