Equivalent ε-definition of singural measures

Question

Sorry in advance for any mistakes in terminology or grammar. In greek we have quite different terminology.

If $μ,ν $ are two measures in measure space $(X, \mathcal{A})$, prove that $μ \botν$ (singular measures) iff $\forall ε>0$ $ \exists A\in\mathcal{A} $ such that $μ(A)<ε$ and $ν(X\backslash A)<ε.$

Correct me if I am wrong but $(\Rightarrow)$ is easy (from the definition of singular measures). I tried this for $(\Leftarrow)$: found a sequence of sets $B_n$ with measure $μ( B_n)<1/2^n $ and $ν(Χ \backslash B_n)<1/2^n $ and I proved that $μ(\bigcap B_n)$ and $ ν(X\backslash\bigcup B_n)$ converge to 0. I can't go any further. Any suggestions?

Answer 1

Let $B=\lim \sup B_n$ (the set of points that belong to $B_n$ for infinitely many $n$. Then $\mu (B)=0$. [$\mu (B) =\mu (\bigcap_n \bigcup_{m \geq n} B_n)\leq \sum\limits_{m=n}^{\infty} \mu( B_n) \to 0$ since $\sum \frac 1 {2^{n}} <\infty$].

Also $$\nu (\bigcap_{m \geq n} (X\setminus B_m))\leq \nu (X\setminus B_n) \to 0$$ so $$\nu (\bigcup_n (\bigcap_{m \geq n} (X\setminus B_m)))=0.$$ This means $\nu (X\setminus B)=0$.