# Equivalent ε-definition of singural measures

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?

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 $$\nu (X\setminus B)=0$$.
• Why is $ν(\bigcup _{m \geq n}(X\backslash B_m))\leq ν(X\backslash B_n)\rightarrow 0$? Do you mean $\bigcap$ or $\sum ν(X\backslash B_n)$? Aug 25, 2021 at 8:43