# Absolute continuity (measures): finiteness

• Oct 31st 2012, 04:28 AM
huberscher
Absolute continuity (measures): finiteness
Hi there

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

If $\mu$ is finite then $\lambda$ 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:

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

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

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.

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

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

Regards
• Oct 31st 2012, 04:42 AM
girdav
Re: Absolute continuity (measures): finiteness
Are you working with non-negative measures?
• Oct 31st 2012, 06:58 AM
huberscher
Re: Absolute continuity (measures): finiteness
yes regular measures non-negative ones.

what do you think? is the statement true?
• Oct 31st 2012, 07:16 AM
girdav
Re: Absolute continuity (measures): finiteness
Let $\nu(A):=\int_Ae^{-x^2}d\lambda(x)$, where $\lambda$ is the Lebesgue measure. $\nu$ is a finite measure, and if $\nu(A)=0$, then $\lambda(A\cap [n,n+1])=0$ for all integer $n$.
• Oct 31st 2012, 08:00 AM
huberscher
Re: Absolute continuity (measures): finiteness
Thank you. But if I define $\lambda$ to be the Lebesgue measure restricted to $\mathbb{R}_{\ge 0}$.

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

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

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

Is this enough? What do you think?

Regards.
• Oct 31st 2012, 08:06 AM
girdav
Re: Absolute continuity (measures): finiteness
What are the properties of Lebesgue measure you use?

If $\nu(A)=0$, then $0=\nu_{A\cap [n,n+1]}=\int_{A\cap (n,n+1)}e^{-x^2}dx\geq \exp(-(n+1)^2)\lambda(A\cap [n,n+1])$.
• Oct 31st 2012, 08:17 AM
huberscher
Re: Absolute continuity (measures): finiteness
In your example I don't understand the whole relation now because $\lambda(A) \neq \lambda(A\cap[n,n+1])$ ??
Hasn't the $\nu$ to be made this way:
$\nu(A)=\int_A e^{-x^2}*d\lambda(A \cap [n,n+1])$
So this n can be fixed?

Concerning my example:

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

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

So my example works as well, does it?
• Oct 31st 2012, 08:21 AM
girdav
Re: Absolute continuity (measures): finiteness
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 $n$ I show that $\lambda(A\cap [n,n+1])$ is $0$. This proves, by $\sigma$-additivity, that $\lambda(A)=0$.
• Nov 19th 2012, 08:31 PM
wallisonline
Re: Absolute continuity (measures): finiteness
What is the difference between absolute continuity and differential continuity?

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