Absolutely continuous measures

Question

We have two measures with

$\mu_F(]-\infty,x])=F(x)$

$\mu_G(]-\infty,x])=G(x)$

$G(x)=F(x)^2$

Prove that $\mu_G$ is absolutely continuous with respect to $\mu_F$.

So we have to prove that $\mu_F(A)=0\Rightarrow\mu_G(A)=0$. I was able to prove this when A is an interval. But how can I argue that it is true for A that are not intervals as well?

Sorry, in another section I read in the meantime that only finite measures are considered. — milfor, Dec 15 '20 at 21:06

score 1 · Accepted Answer · answered Dec 15 '20 at 23:23

Let $\mu_F(E)=0$. By the Approximation Theorem of Measure Theory (See Theorem D, p. 56 of Measure Theory by Halmos) we can find a finite disjoint union on half-closed intervals, say $H=\bigcup_{k=1}^{n} (a_k,b_k]$ such that $\mu (E\Delta H) <\epsilon$ where $\mu =\mu_F+\mu_G$. Now $$\mu_G(E)$$ $$ \leq \mu_G(H)+\mu_G(E \setminus H)$$ $$ < \sum\limits_{k=1}^{n} (G(b_k)-G(a_k))+\epsilon.$$ But $G(b)-G(a) \leq 2(F(b)-F(a))$ so we get $\mu_G(E) \leq 2\mu_F(H)+\epsilon <\mu_F(E)+2\epsilon=2\epsilon$ Since $\epsilon$ is arbitary we are done.

Absolutely continuous measures

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