[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.[/step]

[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: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]