Absolute continuity (measures): finiteness

Hi there

I don't know how to solve this exercise:

""""

Let be two measures on the borel set with (this means that is absolute continuous related to ). Prove the following statement or the contrary:

If is finite then 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:

and the assumption is

So now:

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.

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 , where is the Lebesgue measure. is a finite measure, and if , then for all integer .

Re: Absolute continuity (measures): finiteness

Thank you. But if I define to be the Lebesgue measure restricted to .

and so this is finite.

But how can I prove that if then ??

Since is zero for singletons and the emptyset only ( is a non-negative even function on ) 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 , then .

Re: Absolute continuity (measures): finiteness

In your example I don't understand the whole relation now because ??

Hasn't the to be made this way:

So this n can be fixed?

Concerning my example:

with (this is because f is an even non-negative function on R)

For the Lebesgue measure restricted to it follows that

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 I show that is . This proves, by -additivity, that .

Re: Absolute continuity (measures): finiteness

What is the difference between absolute continuity and differential continuity?

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