Absolute continuity (measures): finiteness

Oct 2012
48
1
Bonn
Hi there

I don't know how to solve this exercise:
""""
Let \(\displaystyle \mu, \lambda \) be two measures on the borel set with \(\displaystyle \lambda << \mu \) (this means that \(\displaystyle \lambda\) is absolute continuous related to \(\displaystyle \mu\)). Prove the following statement or the contrary:

If \(\displaystyle \mu\) is finite then \(\displaystyle \lambda\) is finite.
""""

I'd guess that the statement is true. What's clear is that according to Radon-Nikodym there exists a measurable f such that:

\(\displaystyle \lambda(A)=\int_A f d\mu \) and the assumption is \(\displaystyle \mu(\mathbb{R})<\infty\)

So now: \(\displaystyle \lambda(\mathbb{R})=\int_{\mathbb{R}} f d\mu=...\)

How can I complete this? Is there an upperbound for f I can take (a real number c)? If yes what theorem says that there is one? Then this proof was easy.

\(\displaystyle \int_{\mathbb{R}} f d\mu \le \int_{\mathbb{R}} c d\mu = c*\mu(\mathbb{R})\)

But I can't find a theorem or a corollary that says that. How can I show this?

Regards
 
Jul 2009
678
241
Rouen, France
Are you working with non-negative measures?
 
Oct 2012
48
1
Bonn
yes regular measures non-negative ones.

what do you think? is the statement true?
 
Jul 2009
678
241
Rouen, France
Let \(\displaystyle \nu(A):=\int_Ae^{-x^2}d\lambda(x)\), where \(\displaystyle \lambda\) is the Lebesgue measure. \(\displaystyle \nu\) is a finite measure, and if \(\displaystyle \nu(A)=0\), then \(\displaystyle \lambda(A\cap [n,n+1])=0\) for all integer \(\displaystyle n\).
 
Oct 2012
48
1
Bonn
Thank you. But if I define \(\displaystyle \lambda\) to be the Lebesgue measure restricted to \(\displaystyle \mathbb{R}_{\ge 0}\).

\(\displaystyle \nu(A)=\int_A e^{-x^2} d \lambda \) and \(\displaystyle \nu(\mathbb{R})=\frac{\sqrt{\pi}}{2}\) so this is finite.


But how can I prove that if \(\displaystyle \nu(A)=0 \) then \(\displaystyle \lambda(A)=0\) ??

Since \(\displaystyle \nu\) is zero for singletons and the emptyset only (\(\displaystyle e^{-x^2} \) is a non-negative even function on \(\displaystyle \mathbb{R}_{\ge 0}\)) it follows because of the properties of the Lebesgue measure?

Is this enough? What do you think?

Regards.
 
Last edited:
Jul 2009
678
241
Rouen, France
What are the properties of Lebesgue measure you use?

If \(\displaystyle \nu(A)=0\), then \(\displaystyle 0=\nu_{A\cap [n,n+1]}=\int_{A\cap (n,n+1)}e^{-x^2}dx\geq \exp(-(n+1)^2)\lambda(A\cap [n,n+1])\).
 
Oct 2012
48
1
Bonn
In your example I don't understand the whole relation now because \(\displaystyle \lambda(A) \neq \lambda(A\cap[n,n+1])\) ??
Hasn't the \(\displaystyle \nu\) to be made this way:
\(\displaystyle \nu(A)=\int_A e^{-x^2}*d\lambda(A \cap [n,n+1])\)
So this n can be fixed?


Concerning my example:

\(\displaystyle \nu(A)=0 \Leftrightarrow A=\emptyset \lor A=\{x\} \) with \(\displaystyle x \in \mathbb{R}\) (this is because f is an even non-negative function on R)

For the Lebesgue measure \(\displaystyle \lambda\) restricted to \(\displaystyle \mathbb{R}_{\ge 0}\) it follows that
\(\displaystyle \lambda(\emptyset)=0 \land \lambda(\{x\})=0 \forall x \in \mathbb{R}\)

So my example works as well, does it?
 
Jul 2009
678
241
Rouen, France
Not only countable of finite subsets of the real line have 0 measure. For example, Cantor set is uncountable, but has zero measure.

In my argument, for a fixed \(\displaystyle n\) I show that \(\displaystyle \lambda(A\cap [n,n+1])\) is \(\displaystyle 0\). This proves, by \(\displaystyle \sigma\)-additivity, that \(\displaystyle \lambda(A)=0\).
 
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