[proofplan]
We construct the signed measure $\mu = \mu_1 - \mu_2$ by dominating both $\mu_1$ and $\mu_2$ by the finite measure $\nu = \mu_1 + \mu_2$, applying the Radon-Nikodym theorem to extract densities, and defining $\mu$ via the difference of these densities. The Jordan decomposition of the resulting density provides the Hahn decomposition of $\mu$, confirming it is a well-defined signed measure.
[/proofplan]
[step:Construct a dominating measure and extract Radon-Nikodym densities]
Define the finite positive measure $\nu := \mu_1 + \mu_2$ on $([0,T], \mathcal{B}([0,T]))$. Since $\mu_1$ and $\mu_2$ are both finite Borel measures, $\nu$ is a finite positive measure with $\nu([0,T]) = \mu_1([0,T]) + \mu_2([0,T]) < \infty$.
Both $\mu_1$ and $\mu_2$ are absolutely continuous with respect to $\nu$: for any Borel set $E \subseteq [0,T]$, if $\nu(E) = 0$, then $\mu_1(E) + \mu_2(E) = 0$. Since $\mu_1(E) \geq 0$ and $\mu_2(E) \geq 0$, this forces $\mu_1(E) = \mu_2(E) = 0$. Hence $\mu_1 \ll \nu$ and $\mu_2 \ll \nu$.
By the Radon-Nikodym Theorem, there exist $\nu$-measurable functions
\begin{align*}
f_1: [0,T] &\to [0, \infty), \\
f_2: [0,T] &\to [0, \infty),
\end{align*}
with $f_1, f_2 \in L^1([0,T], \mathcal{B}([0,T]), \nu)$, such that for every Borel set $A \subseteq [0,T]$:
\begin{align*}
\mu_1(A) = \int_A f_1 \, d\nu, \qquad \mu_2(A) = \int_A f_2 \, d\nu.
\end{align*}
The densities $f_1, f_2$ are non-negative $\nu$-a.e. because $\mu_1, \mu_2$ are positive measures.
[guided]
Why do we introduce the dominating measure $\nu = \mu_1 + \mu_2$ rather than working with $\mu_1$ or $\mu_2$ individually? The Radon-Nikodym theorem requires absolute continuity $\mu_i \ll \nu$. If we tried to use $\mu_1$ as the reference measure, we could not guarantee $\mu_2 \ll \mu_1$. By taking the sum, we automatically dominate both.
Why are the densities non-negative? The Radon-Nikodym density $f_i = d\mu_i / d\nu$ satisfies $\mu_i(A) = \int_A f_i \, d\nu$ for all Borel $A$. If $f_i$ were negative on a set $E$ with $\nu(E) > 0$, then $\mu_i(E) = \int_E f_i \, d\nu < 0$, contradicting $\mu_i(E) \geq 0$. So $f_i \geq 0$ $\nu$-a.e.
Why is $f_i \in L^1(\nu)$? Because $\int_{[0,T]} f_i \, d\nu = \mu_i([0,T]) < \infty$.
[/guided]
[/step]
[step:Define the signed measure $\mu$ via the density $g = f_1 - f_2$]
Define the function $g := f_1 - f_2: [0,T] \to \mathbb{R}$. Since $f_1, f_2 \in L^1([0,T], \nu)$, we have $g \in L^1([0,T], \nu)$ with $\|g\|_{L^1(\nu)} \leq \|f_1\|_{L^1(\nu)} + \|f_2\|_{L^1(\nu)} = \mu_1([0,T]) + \mu_2([0,T]) < \infty$.
Decompose $g$ into its positive and negative parts: $g = g^+ - g^-$, where
\begin{align*}
g^+ := \max\{g, 0\} \geq 0, \qquad g^- := \max\{-g, 0\} \geq 0.
\end{align*}
Both $g^+, g^- \in L^1([0,T], \nu)$ since $g^+ \leq |g|$ and $g^- \leq |g|$. Define two finite positive measures:
\begin{align*}
\mu^+(A) := \int_A g^+ \, d\nu, \qquad \mu^-(A) := \int_A g^- \, d\nu,
\end{align*}
for every Borel set $A \subseteq [0,T]$. Each is a well-defined finite positive measure on $([0,T], \mathcal{B}([0,T]))$ because the integrands are non-negative and $L^1(\nu)$.
Define the signed measure:
\begin{align*}
\mu := \mu^+ - \mu^-.
\end{align*}
This is a signed measure in the sense of the Jordan decomposition: $\mu^+$ and $\mu^-$ are finite positive measures, and $\mu(A) = \mu^+(A) - \mu^-(A)$ for every Borel set $A$.
[/step]
[step:Verify $\mu(A) = \mu_1(A) - \mu_2(A)$ for every Borel set $A$]
For any Borel set $A \subseteq [0,T]$:
\begin{align*}
\mu(A) = \mu^+(A) - \mu^-(A) = \int_A g^+ \, d\nu - \int_A g^- \, d\nu = \int_A (g^+ - g^-) \, d\nu = \int_A g \, d\nu,
\end{align*}
where we used linearity of the integral and $g = g^+ - g^-$. Substituting $g = f_1 - f_2$:
\begin{align*}
\mu(A) = \int_A (f_1 - f_2) \, d\nu = \int_A f_1 \, d\nu - \int_A f_2 \, d\nu = \mu_1(A) - \mu_2(A).
\end{align*}
This holds for every Borel set $A \subseteq [0,T]$, completing the proof.
[/step]
