Property of absolute continuity of measures and finite variation

So im trying to prove that if $$\mu$$ is of finite variation and $$\mu <<\nu$$, if $$|\lambda(E_k)| \rightarrow 0$$ then $$\mu(E_k) \rightarrow 0$$. My attempt was: let's consider the set $$E_n = \bigcup\limits_{k=n}^{\infty} E_{k}$$ and $$E=\bigcap\limits_{n=1}^{\infty} E_{n}$$ then $$\nu(E)=0$$ hence $$\mu(E)=0$$ and since it's a finite measure $$\mu(E_k)=0$$. I'm not sure if this is totally correct but I don't see another way of doing it, so any help will be appreciated.

• What kind of measure is $\mu$? A real measure, a complex measure? – Kabo Murphy May 12 at 11:36
• it just says that has finite variation so i guess you can think of it has a real measure. – Pedro Santos May 12 at 11:45

If this is false then there exists $$\epsilon >0$$ and sets $$E_n$$ such that $$\nu (E_n) <\frac 1 {2^{n}}$$ but $$|\mu (E_n)| \geq \epsilon$$ for all $$n$$. This also implies $$|\mu| (E_n) \geq \epsilon$$ for all $$n$$ (where $$|\mu|$$ is the total variation measure associated with $$\mu$$). Let $$F_n=\cup_{i\geq n} E_i$$. If $$F=\cap_n F_n$$ then $$\nu (F)\leq \sum_{i \geq n} \nu(E_i)<\sum_{i \geq n} \frac 1 {2^{i}}$$ for each $$n$$ so $$\nu(F)=0$$. But $$|\mu|(F)=\lim |\mu|(F_n)\geq \epsilon$$. This is a contradiction because $$\mu << \nu$$ implies $$|\mu| << \nu$$.