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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.

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

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