Question about absolutely continuous measures

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In summary, the conversation discusses the construction of a measure \lambda that is absolutely continuous with respect to a given sequence of sigma-finite measures v1, v2, v3, ... The attempt at a solution involves creating an infinite weighted measure, but this approach is hindered by the requirement of sigma-finiteness. The conversation then mentions a theorem that states the existence of a probability measure \lambda that satisfies the given conditions, and provides a proof for this theorem. The speaker expresses confusion about how the original measures v1, v2, ... are related to this mu and expresses the need to show that they are all absolutely continuous with respect to it.
  • #1
jvalton1287
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Homework Statement



Suppose we're given some sigma-finite measures v1, v2, v3,...

I want to construct [tex]\lambda[/tex] such that vn is absolutely continuous w.r.t. [tex]\lambda[/tex] for all n.

2. The attempt at a solution

So far, I've tried thinking of making an infinite weighted (weighted by 2^(-n)). The problem is that sigma-finiteness requires that I can't divide vn(E) by the measure of the entire space. So, I need to find a more clever way of doing this.
 
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  • #2
There's a handy theorem:
Given [tex]\sigma[/tex]-finite measure [tex]\mu[/tex], there exists a probability measure [tex]\lambda[/tex] such that
[tex]\mu(A)=0 \mbox{ iff } \lambda(A)=0[/tex]
Proof:
Let [tex]{A_n}[/tex] be a sequence of measurable sets such that [tex]\Bigcup A_n=X[/tex] and [tex]\mu(A)<\infty[/tex]. Set
[tex]f=\sum_{k=1}^\infty 2^{-k}\frac{\chi_{A_k}}{1+\mu(A_k)}[/tex]
From this we can obtain finite measure [tex]\mu_f[/tex]. I leave to you showing it satisfies the hypothesis.
Good luck!
 
  • #3
I don't quite understand this solution. How are all of the original measures v1, v2, ... related to this mu?

I need to show that all of those initial measures are absolutely continuous w.r.t. to mu.
 
  • #4
Thread locked temporarily. This may be a question on a take-home exam.
 

FAQ: Question about absolutely continuous measures

What is an absolutely continuous measure?

An absolutely continuous measure is a type of probability measure that assigns a measure of zero to sets with zero Lebesgue measure. This means that the measure of any subset of a set with measure zero is also zero.

How is an absolutely continuous measure different from a continuous measure?

An absolutely continuous measure is a type of continuous measure, but not all continuous measures are absolutely continuous. A continuous measure is one that assigns a non-negative measure to all open sets, while an absolutely continuous measure has the additional property of assigning zero measure to sets with zero Lebesgue measure.

What is the importance of absolutely continuous measures in probability theory?

Absolutely continuous measures play a crucial role in probability theory because they allow for the definition of continuous random variables and the calculation of probabilities for continuous distributions. They also help to define important concepts such as probability density functions and cumulative distribution functions.

How are absolutely continuous measures related to Lebesgue integration?

Absolutely continuous measures are a key part of Lebesgue integration, which is a more general and powerful form of integration than Riemann integration. In Lebesgue integration, absolutely continuous measures are used to define the integral of a function over a set, and they allow for the integration of more complex and irregular functions.

Can an absolutely continuous measure be decomposed into simpler measures?

Yes, an absolutely continuous measure can be decomposed into simpler measures through the Lebesgue-Radon-Nikodym theorem. This theorem states that any absolutely continuous measure can be decomposed into a sum of a singular measure and a continuous measure. This decomposition is useful in analyzing and understanding the properties of absolutely continuous measures.

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