[proofplan]
We prove both implications directly from the definition of total variation. If $|\nu|\ll\mu$, then every $\mu$-null set has zero $|\nu|$-mass, and the one-set partition gives $|\nu(A)|\leq |\nu|(A)$, hence $\nu(A)=0$. Conversely, if $\nu\ll\mu$ and $A$ is $\mu$-null, then every measurable subset of $A$ is also $\mu$-null, so $\nu$ vanishes on every piece of every finite measurable partition of $A$; taking the supremum over such partitions gives $|\nu|(A)=0$.
[/proofplan]
[step:Recall the partition formula for total variation]
For $A\in\mathcal E$, let $\Pi(A)$ denote the set of all finite measurable partitions of $A$, meaning all finite families $\mathcal P=\{A_1,\dots,A_n\}$ with $n\in\mathbb N$, each $A_i\in\mathcal E$, the sets $A_i$ pairwise disjoint, and
\begin{align*}
A=\bigcup_{i=1}^n A_i.
\end{align*}
By definition of the total variation measure, for every $A\in\mathcal E$,
\begin{align*}
|\nu|(A)=\sup_{\mathcal P\in\Pi(A)}\sum_{B\in\mathcal P}|\nu(B)|.
\end{align*}
In particular, since $\{A\}\in\Pi(A)$ whenever $A\in\mathcal E$, we have
\begin{align*}
|\nu(A)|\leq |\nu|(A).
\end{align*}
[/step]
[step:Use domination by total variation to prove $|\nu|\ll\mu$ implies $\nu\ll\mu$]
Assume $|\nu|\ll\mu$. Let $A\in\mathcal E$ satisfy $\mu(A)=0$. By [absolute continuity](/page/Absolute%20Continuity) of $|\nu|$ with respect to $\mu$, we have $|\nu|(A)=0$. From the inequality $|\nu(A)|\leq |\nu|(A)$, it follows that
\begin{align*}
|\nu(A)|\leq 0.
\end{align*}
Since $|\nu(A)|\geq 0$, we obtain $|\nu(A)|=0$, and therefore $\nu(A)=0$. Hence $\nu\ll\mu$.
[/step]
[step:Show that $\nu\ll\mu$ forces total variation to vanish on every $\mu$-null set]
Assume $\nu\ll\mu$. Let $A\in\mathcal E$ satisfy $\mu(A)=0$. If $B\in\mathcal E$ and $B\subseteq A$, then positivity of $\mu$ gives
\begin{align*}
0\leq \mu(B)\leq \mu(A)=0.
\end{align*}
Thus $\mu(B)=0$, and by $\nu\ll\mu$ we have $\nu(B)=0$.
Now let $\mathcal P=\{A_1,\dots,A_n\}\in\Pi(A)$ be an arbitrary finite measurable partition of $A$. For each $i\in\{1,\dots,n\}$, the set $A_i$ is measurable and satisfies $A_i\subseteq A$, so $\nu(A_i)=0$. Therefore
\begin{align*}
\sum_{B\in\mathcal P}|\nu(B)|=0.
\end{align*}
Taking the supremum over all $\mathcal P\in\Pi(A)$ yields
\begin{align*}
|\nu|(A)=0.
\end{align*}
Since every $A\in\mathcal E$ with $\mu(A)=0$ satisfies $|\nu|(A)=0$, we conclude that $|\nu|\ll\mu$.
[guided]
Assume $\nu\ll\mu$. To prove $|\nu|\ll\mu$, we must start with an arbitrary $\mu$-null measurable set and prove that its total variation is zero. Let $A\in\mathcal E$ satisfy $\mu(A)=0$.
The definition of $|\nu|(A)$ involves all finite measurable partitions of $A$, so we need to understand what $\nu$ does on every measurable piece of $A$. Let $B\in\mathcal E$ be any measurable subset of $A$. Since $\mu$ is a positive measure, monotonicity gives
\begin{align*}
0\leq \mu(B)\leq \mu(A)=0.
\end{align*}
Hence $\mu(B)=0$. The hypothesis $\nu\ll\mu$ means exactly that $\mu$-null measurable sets are $\nu$-null, so $\nu(B)=0$.
Now take an arbitrary finite measurable partition $\mathcal P=\{A_1,\dots,A_n\}\in\Pi(A)$. Each set $A_i$ is measurable and satisfies $A_i\subseteq A$. Applying the conclusion from the previous paragraph with $B=A_i$, we get $\nu(A_i)=0$ for every $i\in\{1,\dots,n\}$. Therefore each summand in the partition sum defining total variation is zero, and hence
\begin{align*}
\sum_{B\in\mathcal P}|\nu(B)|=0.
\end{align*}
Because this holds for every finite measurable partition $\mathcal P$ of $A$, the supremum of all these partition sums is also zero. By the definition of total variation,
\begin{align*}
|\nu|(A)=\sup_{\mathcal P\in\Pi(A)}\sum_{B\in\mathcal P}|\nu(B)|=0.
\end{align*}
Thus every $\mu$-null measurable set is also $|\nu|$-null, which is precisely $|\nu|\ll\mu$.
[/guided]
[/step]
[step:Combine the two implications]
The second step proved that $|\nu|\ll\mu$ implies $\nu\ll\mu$, and the third step proved that $\nu\ll\mu$ implies $|\nu|\ll\mu$. Therefore
\begin{align*}
\nu\ll\mu \iff |\nu|\ll\mu.
\end{align*}
[/step]
