[proofplan]
We first pass from [absolute continuity](/page/Absolute%20Continuity) of the complex measure $\nu$ to absolute continuity of its total variation measure $|\nu|$, using [citetheorem:10124]. The forward implication is proved by contradiction: if no $\delta$ works for some $\varepsilon>0$, we choose sets $A_n$ with rapidly summable $\mu$-measure but uniformly positive $|\nu|$-measure. Tail unions then have $\mu$-measure tending to zero, while continuity from above for the finite measure $|\nu|$ forces their $|\nu|$-measure to tend to zero, contradicting the lower bound. The reverse implication is immediate from the epsilon-delta condition applied to $\mu$-null sets.
[/proofplan]
[step:Pass absolute continuity from $\nu$ to its total variation measure]
Assume $\nu\ll\mu$. Since $\nu:\mathcal E\to\mathbb C$ is a finite complex measure and $\mu$ is a positive measure, [citetheorem:10124] applies and gives
\begin{align*}
|\nu|\ll\mu.
\end{align*}
Thus, for every $N\in\mathcal E$,
\begin{align*}
\mu(N)=0 \implies |\nu|(N)=0.
\end{align*}
[guided]
Assume $\nu\ll\mu$. The epsilon-delta condition in the theorem is stated in terms of the positive measure $|\nu|$, not directly in terms of the complex measure $\nu$. Therefore the first task is to transfer absolute continuity from $\nu$ to its total variation measure.
The hypotheses of [citetheorem:10124] are exactly present here: $(E,\mathcal E)$ is a measurable space, $\mu$ is a positive measure, and $\nu:\mathcal E\to\mathbb C$ is a complex measure. Hence that result gives
\begin{align*}
\nu\ll\mu \iff |\nu|\ll\mu.
\end{align*}
Since we are assuming $\nu\ll\mu$, we obtain
\begin{align*}
|\nu|\ll\mu.
\end{align*}
Unpacking this conclusion, for every measurable set $N\in\mathcal E$,
\begin{align*}
\mu(N)=0 \implies |\nu|(N)=0.
\end{align*}
This is the form of absolute continuity needed for the contradiction argument, because the contradiction will produce a measurable null set for $\mu$ that must also be null for $|\nu|$.
[/guided]
[/step]
[step:Construct small $\mu$-sets with uniformly large $|\nu|$-measure if the condition fails]
Suppose, toward a contradiction, that the epsilon-delta condition fails. Then there exists $\varepsilon_0>0$ such that for every $\delta>0$ there is a set $A\in\mathcal E$ satisfying
\begin{align*}
\mu(A)<\delta
\end{align*}
and
\begin{align*}
|\nu|(A)\ge\varepsilon_0.
\end{align*}
For each $n\in\mathbb N$, apply this failure with $\delta=2^{-n}$ and choose $A_n\in\mathcal E$ such that
\begin{align*}
\mu(A_n)<2^{-n}
\end{align*}
and
\begin{align*}
|\nu|(A_n)\ge\varepsilon_0.
\end{align*}
[/step]
[step:Use tail unions to obtain a $\mu$-null limiting set]
For each $n\in\mathbb N$, define the tail union
\begin{align*}
B_n:=\bigcup_{k=n}^{\infty} A_k.
\end{align*}
Then $B_n\in\mathcal E$, $A_n\subset B_n$, and $B_{n+1}\subset B_n$ for every $n\in\mathbb N$. By [countable subadditivity](/theorems/1108) of the positive measure $\mu$,
\begin{align*}
\mu(B_n)\le \sum_{k=n}^{\infty}\mu(A_k).
\end{align*}
Using $\mu(A_k)<2^{-k}$ for every $k\in\mathbb N$, we get
\begin{align*}
\mu(B_n)\le \sum_{k=n}^{\infty}2^{-k}=2^{1-n}.
\end{align*}
Define
\begin{align*}
B:=\bigcap_{n=1}^{\infty}B_n.
\end{align*}
Since $B\subset B_n$ for every $n\in\mathbb N$, monotonicity of $\mu$ gives
\begin{align*}
\mu(B)\le 2^{1-n}
\end{align*}
for every $n\in\mathbb N$. Letting $n\to\infty$ yields $\mu(B)=0$.
[guided]
The sets $A_n$ are individually small for $\mu$, but they are not nested. To use continuity from above later, we replace them by nested tail unions. For each $n\in\mathbb N$, define
\begin{align*}
B_n:=\bigcup_{k=n}^{\infty} A_k.
\end{align*}
Each $B_n$ is measurable because $\mathcal E$ is closed under countable unions. Also $A_n\subset B_n$, and the sequence is decreasing because removing the first set from a tail can only shrink the union:
\begin{align*}
B_{n+1}\subset B_n.
\end{align*}
We now estimate the $\mu$-measure of $B_n$. Countable subadditivity of the positive measure $\mu$ gives
\begin{align*}
\mu(B_n)\le \sum_{k=n}^{\infty}\mu(A_k).
\end{align*}
By construction, $\mu(A_k)<2^{-k}$ for every $k\in\mathbb N$, so
\begin{align*}
\mu(B_n)\le \sum_{k=n}^{\infty}2^{-k}=2^{1-n}.
\end{align*}
The geometric tail tends to zero as $n\to\infty$.
Now define the limiting set
\begin{align*}
B:=\bigcap_{n=1}^{\infty}B_n.
\end{align*}
This set is measurable because $\mathcal E$ is closed under countable intersections. Since $B\subset B_n$ for every $n\in\mathbb N$, monotonicity of $\mu$ gives
\begin{align*}
\mu(B)\le \mu(B_n)\le 2^{1-n}
\end{align*}
for every $n\in\mathbb N$. The only nonnegative number bounded above by $2^{1-n}$ for all $n$ is $0$, hence
\begin{align*}
\mu(B)=0.
\end{align*}
This produces the $\mu$-null set on which the absolute continuity of $|\nu|$ will be used.
[/guided]
[/step]
[step:Apply continuity from above to contradict the uniform lower bound]
Since $|\nu|\ll\mu$ and $\mu(B)=0$, we have
\begin{align*}
|\nu|(B)=0.
\end{align*}
The measure $|\nu|$ is finite because $\nu$ is a finite complex measure. Since $(B_n)_{n\in\mathbb N}$ is a decreasing sequence in $\mathcal E$ with intersection $B$, continuity from above for the finite positive measure $|\nu|$ gives
\begin{align*}
\lim_{n\to\infty}|\nu|(B_n)=|\nu|(B)=0.
\end{align*}
On the other hand, $A_n\subset B_n$, so monotonicity of $|\nu|$ gives
\begin{align*}
|\nu|(B_n)\ge |\nu|(A_n)\ge\varepsilon_0
\end{align*}
for every $n\in\mathbb N$. This contradicts the convergence $|\nu|(B_n)\to0$. Therefore the epsilon-delta condition holds whenever $\nu\ll\mu$.
[/step]
[step:Recover absolute continuity of $\nu$ from the epsilon-delta condition]
Conversely, assume that for every $\varepsilon>0$ there exists $\delta>0$ such that, for every $A\in\mathcal E$,
\begin{align*}
\mu(A)<\delta \implies |\nu|(A)<\varepsilon.
\end{align*}
Let $N\in\mathcal E$ satisfy $\mu(N)=0$. Fix $\varepsilon>0$, and choose the corresponding $\delta>0$. Since $\mu(N)=0<\delta$, the hypothesis gives
\begin{align*}
|\nu|(N)<\varepsilon.
\end{align*}
Because this holds for every $\varepsilon>0$ and $|\nu|(N)\ge0$, we have $|\nu|(N)=0$. By the defining domination of a complex measure by its total variation,
\begin{align*}
|\nu(N)|\le |\nu|(N)=0,
\end{align*}
so $\nu(N)=0$. Hence $\nu\ll\mu$, completing the proof.
[/step]
