# Absolute continuity (measures): finiteness

#### huberscher

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

#### girdav

Are you working with non-negative measures?

#### huberscher

yes regular measures non-negative ones.

what do you think? is the statement true?

#### girdav

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 $$\displaystyle n$$.

#### huberscher

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.

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#### girdav

What are the properties of Lebesgue measure you use?

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

#### huberscher

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?

#### girdav

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 $$\displaystyle \lambda(A)=0$$.

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