Are you working with non-negative measures?
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
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?
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?
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?
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 .