I am having difficulty solving an exercise in the book "Modern Real Analysis" by Ziemer. It is exercise 7 of section 6 on signed measures in the chapter on integration (chapter 6). The exercise statement is :
Let $(X,\mathcal{M},\mu)$ be a finite measure space with $\mu(X) < \infty$. Let $\nu_{k}$ be asequence of finite measures on $\mathcal{M}$ (that is, $\nu_{k}(X) < \infty$ for all $k$) with theproperty that they are uniformly absolutely continuous with respect to $\mu$; that is,for each $\epsilon > 0$, there exist $\delta > 0$ and a positive integer $K$ such that$\nu_{k}(E) < \epsilon$ for all $k \geq K$ and all $E \in \mathcal{M}$ for which $\mu(E) < \delta$.Assume that the limit\begin{equation}\nu(E) := \lim_{k \rightarrow \infty} \nu_{k}(E) \text{ , } E \in \mathcal{M}\end{equation}exists. Prove that $\nu$ is a $\sigma$-finite measure on $\mathcal{M}$.
Here is my attempt so far :
Let's suppose $\nu$ is not $\sigma$-finite. Then $\{ E_{i} \}_{i \in \mathbb{N}}$ such that each $E_{i} \in \mathcal{M}$,$E_{i} \cap E_{j} = \emptyset$ when $i \neq j$, $\nu(E_{i}) < \infty$ for each $i \in \mathbb{N}$, and $X = \bigcup_{i \in \mathbb{N}} E_{i}$implies :\begin{equation} \tag{1}\label{1}\sum_{i \in \mathbb{N}} \nu(E_{i}) = \infty\end{equation}Let $\nu_{k}(X) = \alpha_{k} < \infty$. So :\begin{equation} \tag{2}\label{2}\sum_{i \in \mathbb{N}} \nu_{k}(E_{i}) = \alpha_{k}\end{equation}Choose $\epsilon > 0$. Then $\exists \delta > 0$ and $K \in \mathbb{N}$ such that\begin{equation} \tag{3}\label{3}\mu(E_{i}) < \delta \Rightarrow \nu_{k}(E_{i}) < \epsilon \; \forall k \geq K \end{equation}for each $i \in \mathbb{N}$. So $\nu_{k} \rightarrow \nu$ implies :\begin{equation} \tag{4}\label{4}\mu(E_{i}) < \delta \Rightarrow \nu(E_{i}) \leq \epsilon\end{equation}for each $i \in \mathbb{N}$. So if $\mu(E_{i}) < \delta \; \forall i \in \mathbb{N}$then \eqref{2}, \eqref{3}, and \eqref{4} implies $\sum_{i \in \mathbb{N}} \nu(E_{i}) < \infty$ and $\eqref{1}$ is contradicted.
We only need to show that there exists at least one partition such that \eqref{1} fails tobe true. We need a partition such that\begin{equation*}\mu(E_{i}) < \delta \; \forall i \in \mathbb{N}\end{equation*}If $\mu$ is a non-atomic measure (Wikipedia) then :\begin{equation*}A \in \mathcal{M} \text{ and } \mu(A) > 0 \Rightarrow B \subset A \text{ such that } B \in \mathcal{M} \text{ and } \mu(A) > \mu(B) > 0 \end{equation*}So if $\mu$ is non-atomic we can find $\{ E_{i} \}_{i \in \mathbb{N}}$ so that $\mu(E_{i}) < \delta \; \forall i \in \mathbb{N}$ and\begin{equation*}\eqref{2} \text{ and } \eqref{3} \text{ and } \eqref{4} \Rightarrow \neg \eqref{1}\end{equation*}and $\nu$ is $\sigma$-finite. $\checkmark$
I'm not sure how to prove the case when $\mu$ is not non-atomic.