I don't know how to solve this exercise:

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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.

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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