# Showing that $\nu \ll \mu$ implies $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\mu(A) < \delta \implies \nu(A) < \epsilon$

I am stuck on what I think may be the very last line of the proof I am seeking.

Let $(X, \mathcal{B})$ be a measurable space which has associated with it the finite measures $\mu$ and $\nu$ s.t. $\nu \ll \mu$. I aim to show that $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\forall A \in \mathcal{B}$,

$$\mu(A) < \delta \implies \nu(A) < \epsilon$$

1. Fix $\epsilon > 0$.

2. For all $n \in \mathbb{N}$, let $\delta_n = \frac{1}{n^2}$.

3. For all $n \in \mathbb{N}$, let $A_n \in \mathcal{B}$ s.t. if $\exists E \in \mathcal{B}$ s.t. $\mu(E) < \delta_n$ and $\nu(E) > \epsilon$, then set $A_n = E$. Otherwise, set $A_n = \emptyset \in \mathcal{B}$.

4. Suppose for sake of contradiction that $|\{A_n\}| = \infty$, so that no matter the $\delta > 0$, we could find a $\delta_n = \frac{1}{n^2} < \delta$ which has associated with it a measurable $A_n \ne \emptyset$ with $\mu(A_n) < \delta_n < \delta$ and $\nu(A_n) > \epsilon$.

5. Now if we let $\underset{n \rightarrow \infty}{\text{limsup}}$ $A_n = S$, we have (from a prior problem) that $\mu(S) = 0$ since $\mu$ is a finite measure and $\sum_{n=1}^\infty \mu(A_n) \le \sum_{n=1}^\infty \delta_n = \sum_{n=1}^\infty \frac{1}{n^2} < \infty$. Since $\nu \ll \mu$ we therefore have $\nu(S) = 0$ as well.

6. Yet $\nu(S) = \nu(\bigcap_{n=1}^\infty \bigcup_{n=m}^\infty A_m) \ge \epsilon > 0$ since...

and it's here where I'm stuck in the proof.

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What do you mean by "finite measurable space"? I only know a definition for the space $(X, \mathcal B, \mu)$ to be finite, but not for $(X,\mathcal B)$ alone. –  Patrick Da Silva Dec 18 '12 at 23:52
Should have read "measurable space" and not "finite measurable space". Fixed above. –  user1770201 Dec 18 '12 at 23:54
Yes, that is more comprehensible now. Thank you –  Patrick Da Silva Dec 18 '12 at 23:55
What is you definition of $\nu \ll \mu$? –  leo Dec 19 '12 at 20:41

You are almost there. Notice that since $\nu(S) = 0$, $$0 = \nu(S) = \nu \left( \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} A_m \right) = \lim_{n \to \infty} \nu \left( \bigcup_{m=n}^{\infty} A_m \right).$$ Therefore, there exists $n$ such that $$\nu(A_n) \le \nu \left( \bigcup_{m=n}^{\infty} A_m \right) < \varepsilon,$$ a contradiction of the choice of $A_n$.
I must say though that the right way to formulate this proof would be this : Suppose by sake of contradiction that the result is false, i.e. that there exists an $\varepsilon > 0$ such that for all $\delta > 0$, there is $C_{\delta} \in \mathcal B$ with $$\mu(C_{\delta}) < \delta, \quad \nu(C_{\delta}) \ge \varepsilon.$$ Then you can choose $\delta_n = \frac 1{n^2}$ and $A_n = C_{\delta_n}$ and then continue by following our steps.
@user1770201 : I am just saying that you look like you're not comfortable with the $\varepsilon-\delta$ proofs because you feel like putting a lot of unnecessary comments. Usually, when we do a proof by contradiction with those kind of statements, it's better to suppose the whole thing is wrong instead of starting the contradiction hypothesis in the middle. –  Patrick Da Silva Dec 19 '12 at 3:37