Absolute continuity (measures): finiteness

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

Re: Absolute continuity (measures): finiteness

Are you working with non-negative measures?

Re: Absolute continuity (measures): finiteness

yes regular measures non-negative ones.

what do you think? is the statement true?

Re: Absolute continuity (measures): finiteness

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 $\displaystyle n$.

Re: Absolute continuity (measures): finiteness

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.

Re: Absolute continuity (measures): finiteness

What are the properties of Lebesgue measure you use?

If $\displaystyle \nu(A)=0$, then $\displaystyle 0=\nu_{A\cap [n,n+1]}=\int_{A\cap (n,n+1)}e^{-x^2}dx\geq \exp(-(n+1)^2)\lambda(A\cap [n,n+1])$.

Re: Absolute continuity (measures): finiteness

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?

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 $\displaystyle n$ I show that $\displaystyle \lambda(A\cap [n,n+1])$ is $\displaystyle 0$. This proves, by $\displaystyle \sigma$-additivity, that $\displaystyle \lambda(A)=0$.

Re: Absolute continuity (measures): finiteness

What is the difference between absolute continuity and differential continuity?

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