[proof]
**Proof of Uniqueness**
Suppose we have two such decompositions of $\nu$:
\begin{align*}
\nu = \nu_{ac} + \nu_s \quad \text{and} \quad \nu = \nu'_{ac} + \nu'_s
\end{align*}
where $\nu_{ac} \ll \mu$, $\nu'_{ac} \ll \mu$, $\nu_s \perp \mu$, and $\nu'_s \perp \mu$. Our goal is to show that $\nu_{ac} = \nu'_{ac}$ and $\nu_s = \nu'_s$.
**Step 1 (Equate the decompositions)**
From the two decompositions, we can write:
\begin{align*}
\nu_{ac} + \nu_s = \nu'_{ac} + \nu'_s
\end{align*}
Rearranging this equation gives:
\begin{align*}
\nu_{ac} - \nu'_{ac} = \nu'_s - \nu_s
\end{align*}
Let's define a signed measure $\lambda = \nu_{ac} - \nu'_{ac}$. Our task is to show that $\lambda$ is the zero measure.
**Step 2 (Show that λ is absolutely continuous with respect to μ)**
Let $A$ be a Borel set such that $\mu(A) = 0$. Since $\nu_{ac} \ll \mu$, we have $\nu_{ac}(A) = 0$. Similarly, since $\nu'_{ac} \ll \mu$, we have $\nu'_{ac}(A) = 0$. The Jordan decomposition for signed measures allows us to state that if the positive and negative parts of the measure are zero on A, then the measure of A is zero. Therefore,
\begin{align*}
\lambda(A) = \nu_{ac}(A) - \nu'_{ac}(A) = 0 - 0 = 0.
\end{align*}
This holds for any set $A$ with $\mu(A) = 0$, so we conclude that $\lambda \ll \mu$.
**Step 3 (Show that λ is singular with respect to μ)**
Since $\nu_s \perp \mu$, there exists a set $B_1$ such that $\mu(B_1)=0$ and $\nu_s(\mathbb{R}^n \setminus B_1)=0$.
Since $\nu'_s \perp \mu$, there exists a set $B_2$ such that $\mu(B_2)=0$ and $\nu'_s(\mathbb{R}^n \setminus B_2)=0$.
Let $B = B_1 \cup B_2$. By the sub-additivity of measures, $\mu(B) \le \mu(B_1) + \mu(B_2) = 0+0=0$. So, $\mu(B)=0$.
Now consider the complement, $\mathbb{R}^n \setminus B = (\mathbb{R}^n \setminus B_1) \cap (\mathbb{R}^n \setminus B_2)$.
For any measurable set $E \subseteq \mathbb{R}^n \setminus B$, we have $E \subseteq \mathbb{R}^n \setminus B_1$, which implies $\nu_s(E) = 0$. Similarly, $E \subseteq \mathbb{R}^n \setminus B_2$, which implies $\nu'_s(E)=0$. This shows that both $\nu_s$ and $\nu'_s$ are concentrated on $B$. Therefore, their difference, $\lambda = \nu'_s - \nu_s$, must also be concentrated on $B$. Formally, for any set $E \subseteq \mathbb{R}^n \setminus B$, $\lambda(E) = \nu'_s(E) - \nu_s(E) = 0-0=0$. This means $\lambda(\mathbb{R}^n \setminus B)=0$.
Since we have found a set $B$ with $\mu(B) = 0$ and $\lambda(\mathbb{R}^n \setminus B) = 0$, we conclude that $\lambda \perp \mu$.
**Step 4 (Conclude λ is the zero measure)**
We have shown that the signed measure $\lambda$ is both absolutely continuous with respect to $\mu$ and singular with respect to $\mu$.
Since $\lambda \perp \mu$, there is a set $B$ with $\mu(B) = 0$ and $\lambda(\mathbb{R}^n \setminus B) = 0$.
Since $\lambda \ll \mu$ and $\mu(B) = 0$, we must have $\lambda(B) = 0$.
The total measure of the entire space is
\begin{align*}
\lambda(\mathbb{R}^n) = \lambda(B) + \lambda(\mathbb{R}^n \setminus B) = 0 + 0 = 0.
\end{align*}
A similar argument for the positive and negative parts in the Jordan decomposition of $\lambda$ shows that $\lambda$ must be the zero measure. That is, $\lambda(A) = 0$ for all Borel sets $A$.
**Step 5 (Establish uniqueness)**
Since $\lambda = 0$, we have $\nu_{ac} - \nu'_{ac} = 0$, which implies $\nu_{ac} = \nu'_{ac}$.
It follows immediately that $\nu'_s - \nu_s = 0$, which implies $\nu_s = \nu'_s$.
This completes the proof of uniqueness.
[/proof]