Here is one problem I am stuck on:
Prove that a measure $\eta$ is absolutely continuous wrt $\mu$ $\longleftrightarrow$ $\forall \epsilon > 0, \exists \delta > 0$ such that $\mu(A) < \delta$ implies that $\eta(A) < \epsilon$.
This result makes sense to me intuitively because it matches with my notion of continuity from real analysis. But I'm having trouble proving this fact.
I've also got some other theorems I can use, like Radon-Nikodym, but I don't think it is necessary here (I could be wrong). I'm pretty sure all you need is the definition, and I haven't been able to make much progress with it. I would really appreciate your help in solving this problem.
It seems like this sort of problem would be a "common" measure theory proof but I can't come up with the right search terms either to find a helpful result (so if you happen to know of one, that would also be helpful).
Suppose $\eta$ is absolutely continuous w.r.t. $\mu$. By considering the positive and negative parts of $\eta$ we can reduce the proof to the case when $\eta$ is a positive finite measure.
Suppose there exists some $\epsilon >0$ such that there is no $\delta$ with the property that $\mu(A) <\delta$ implies $\eta (A) <\epsilon$. Then there is a sequence $(A_n)$ such that $\mu (A_n) <\frac 1 {2^{n}}$ and $\eta (A_n) \geq \epsilon$ for all $n$. Let $A =\lim \sup A_n =\cap_k \cup_{j \geq k} A_j$. Then $\mu(A) \leq \sum\limits_{k=n}^{\infty} \frac 1 {2^{k}}$ for each $n$ which implies $\mu (A)=0$. Hence $\eta (A)=0$. Now $\eta (\lim \inf A_n) \leq \lim inf \eta (A_n)$ by Fatou's Lemma. Taking complements this gives $\eta (A) \geq \lim \sup \eta(A_n) \geq \epsilon >0$ which is a contradiction.
The converse part is trivial: if $\mu(A)=0$ then $\mu (A) <\delta$ so we get $\eta (A) <\epsilon $ for every $\epsilon >0$ which implies $\eta (A)=0$.