Let ( <mi mathvariant="normal">&#x03A9;<!-- Ω --> , <mrow class="MJX-TeXAtom-ORD">

Ayaan Barr 2022-07-03 Answered
Let ( Ω , A ) be a measurable space, and let μ : A [ 0 , + ) be a monotone subadditive set function. Let moroever A 1 , A 2 , A be a sequence of measurable sets. It is easy to prove that μ ( A n ) 0 μ ( A n c ) μ ( Ω ). On the contrary, I am not able neither to prove the opposite implication, namely μ ( A n ) μ ( Ω ) μ ( A n c ) 0, nor to find an example violating it. Can someone find either one or the other?
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)

Sophia Mcdowell
Answered 2022-07-04 Author has 14 answers
It's not true that μ ( A n ) μ ( Ω ) μ ( A n c ) 0. Consider R endowed with the Lebesgue σ-algebra. Let A n = [ 0 , n ]. Then μ ( A n ) = n + = μ ( R ), but μ ( A n c ) = + for all n.

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 2022-08-13
Celsius thermometer Denotes a temperature of 36.6 degrees Celsius. What is the corresponding Fahrenheit reading of the thermometer?
asked 2022-06-24
Pratt's Lemma is : ξ , η , ζ and ξ n , η n , ζ n such that:
ξ n ξ , η n η , ζ n ζ , convergence in probability
and η n ξ n ζ n , E ζ n E ζ , E η n E η, and E ζ , E η , E ξ are finite, prove :
If η n 0 ζ n , then E | ξ n ξ | 0.

I know how to prove E ξ n E ξ, but E | ξ n ξ | 0 seems not easy proved from it.
My first question is how to prove E | ξ n ξ | 0.
And I don't know why should emphasize the condition " η n 0 ζ n ", so my second question is: Is there any example that E | ξ n ξ | 0 is wrong if the condition is not fullfilled.
asked 2022-05-23
MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document I participate the stochastic course and we now speak about summable families. There we have the following definition:

Let Ω be countable and a : Ω R + { } be a map. Then we define
Ω a ( ω ) := sup F Ω , | F | < F a ( ω )

Now our Prof said that we can consider Ω a ( ω ) as the integral of math xmlns="http://www.w3.org/1998/Math/MathML"> a over Ω to get a better connetion to measure theory afterwards when we speak about Fatou's lemma, Beppo Levi theorem ect. Because all this theorems we have seen with integrals last semester.
But somehow I don't see why this sum is equal to the integral. So I know from measure theory that if we have a simple function f : Ω R + { } where f ( Ω ) = { b 1 , . . . , b n } finately many then
Ω f   d μ = i = 1 n f ( b i ) μ ( Ω i )
where Ω i = f 1 ( { b i } ). So but here I don't think that this has to do something with simple functions right?
Therefore I wanted to ask you if someone could explain me why we can see this sum as the integral of a over Ω.
asked 2022-06-26
Let f n : R R be defined by f n ( ω ) = ω ( 1 + 1 n ) and let μ be the Lebesgue measure.
I have to show that f n does not converge in measure μ to ω.
Here's my attempt: By definition, f n converges in measure to ω if:
lim n μ ( ω : | f n ( ω ) ω | δ ) = 0.
Rewriting, it becomes that:
lim n μ ( ω : | ω | n δ ) = 0.
I realized that if δ = 1 for example, then the measure is surely 0 as n . I need to come up with a value of δ such that the measure is non-zero. I thought about letting δ = 1 n ,, but am unsure if it'll become indeterminate once I set n (because I basically get 1 ) .
Does this solution of δ = 1 n truly work as a counterexample? If not, what should I do?
asked 2022-06-27
Suppose that the Hilbert space of a quantum-mechanical system - which we will call the quantum door - is generated by two states, |open> and |closed>, forming an orthonormal basis. Suppose also that the system is prepared in the state
| ψ ( x ) >= 1 5 ( | O P E N > + 2 | C L O S E D > ) We are given a device that measures whether the quantum door is open or closed.
(i)If we perform a measurement, which probability do we have to find the quantum door open?
(ii) Suppose the measurement returns that the quantum door is closed, and assume that the quantum Hamiltonian is identically 0 for this system at any future times. Does the door stay closed forever?
For part (i) I get
P O p e n = ( < o p e n | 1 5 ( | O P E N > + 2 | C L O S E D > ) 2 = 1/5 ?
I also need help with part (ii), i am unure about this.
asked 2022-07-09
I'm studying Pavliotis' Stochastic Processes and I am having trouble with one of the exercises.
Specifically the first exercise of chapter two is prove that the Markov property in the sense off the immediate future being independent of the past given the present, i.e. P ( X n + 1 | X 1 , , X n ) = P ( X n + 1 | X n ) implies that arbitrary futures are independent of the past given the present, i.e. P ( X n + m | X 1 , , X n ) = P ( X n + m | X n ).
Conceptually I imagine a proof using induction along the lines of: Assume the equality holds for some m, then
P ( X n + m + 1 | X 1 , , X n ) = P ( X n + m + 1 , X n + m = x | X 1 , , X n ) d x = P ( X n + m + 1 | X n + m = x ) P ( X n + m = x | X 1 , , X n ) d x = P ( X n + m + 1 | X n + m = x ) P ( X n + m = x | X n ) d x = P ( X n + m + 1 , X n + m = x | X n ) d x = P ( X n + m + 1 | X n )
However, this is based on intuition and my experience non-measure-theoretic probability theory. I cannot justify these steps when I think of P as a measure and P ( X = x ) meaning P { X 1 ( x ) }. In fact I have had trouble finding a definition of conditional probability that is at all helpful. Most authors seem to provide the abstract definition in terms of conditional expectation with respect to σ-algebra, but I haven't found any resources that show how to work with this definition.
So my question is: how (if it all) are these steps, particularly the first two equalities justified from a measure theoretic perspective?
asked 2022-05-20
Is it possible to have lim sup n g h n 1 h n 1 = ?
Consider a metric space ( Ξ , F , ν ) and g and { h n } n non negative, measurable functions in this space. Consider then the norm
g 1 = Ξ | g ( ω ) | d ν ( ω )
and that
1. g 1 <
2. h n 1 < n
3. h n 1, so we have that g h n 1 1 n .
4. lim n h n = 0

My question is, assuming that g doesn't depend on n, is it possible to have
lim sup n g h n 1 h n 1 = ?
The result is true if g and h n were uncorrelated, or at least negativelly correlated (that is, g h n 1 g 1 h n 1 ). But I was wondering if the fact g doesn't depend on n is enough.

New questions