[proofplan] We establish the bijection in two stages. First, we show the forward map $\mu \mapsto a(t) = \mu([0,t])$ is well-defined and lands in the set of cadlag BV functions with $a(0) = 0$: right-continuity follows from continuity of measures from above, left-limits from continuity from below, and the BV property from $V_a(T) \leq |\mu|([0,T])$. Second, we construct the inverse: given a cadlag BV function $a$ with $a(0) = 0$, we decompose $a = a^+ - a^-$ using the Jordan-type decomposition, define positive measures $\mu^\pm$ from the monotone functions $a^\pm$, and verify $\mu = \mu^+ - \mu^-$ recovers $a$. Finally, we check the two maps are inverses and verify the total variation identity. [/proofplan]

[step:Show the forward map $\mu \mapsto a$ produces a cadlag function with $a(0) = 0$]Let $\mu$ be a signed measure on $([0,T], \mathcal{B}([0,T]))$, and write $\mu = \mu^+ - \mu^-$ for its Jordan decomposition into finite positive measures. Define \begin{align*} a: [0,T] &\to \mathbb{R}, \\ t &\mapsto \mu([0,t]). \end{align*} Since $\mu$ is a finite signed measure, $|a(t)| = |\mu([0,t])| \leq |\mu|([0,T]) < \infty$ for all $t \in [0,T]$, so $a$ is well-defined. Moreover, $a(0) = \mu([0,0]) = \mu(\{0\})$. We need $a(0) = 0$, so we adjust the definition: set $a(t) := \mu((0,t])$ for $t > 0$ and $a(0) := 0$. Equivalently, $a(t) = \mu([0,t]) - \mu(\{0\})$. For the remainder of the proof, we use $a(t) = \mu((0,t])$ with $a(0) = 0$. **Right-continuity.** For a decreasing sequence $t_n \downarrow t$ with $t_n > t$, the intervals satisfy $(0, t_n] \downarrow (0, t]$. By continuity of measures from above (valid for finite measures), $\mu((0, t_n]) \to \mu((0, t])$, hence $a(t_n) \to a(t)$. **Existence of left-limits.** For an increasing sequence $t_n \uparrow t$ with $t_n < t$, the intervals satisfy $(0, t_n] \uparrow (0, t)$. By continuity of measures from below, $\mu((0, t_n]) \to \mu((0, t))$. Hence $a(t^-) := \lim_{s \uparrow t} a(s) = \mu((0, t))$ exists and is finite.[/step]

[guided]Why does continuity of measures apply here? A signed measure $\mu = \mu^+ - \mu^-$ with both $\mu^+$ and $\mu^-$ finite inherits continuity from above and below from the corresponding properties of $\mu^+$ and $\mu^-$. For continuity from above: if $A_n \downarrow A$ with $|\mu|(A_1) < \infty$, then $\mu(A_n) = \mu^+(A_n) - \mu^-(A_n) \to \mu^+(A) - \mu^-(A) = \mu(A)$. The finiteness condition $|\mu|(A_1) < \infty$ is satisfied because $|\mu|([0,T]) < \infty$. The left-limit $a(t^-) = \mu((0,t))$ differs from $a(t) = \mu((0,t])$ by exactly $\mu(\{t\})$: the jump of $a$ at $t$ is $\Delta a(t) = a(t) - a(t^-) = \mu(\{t\})$. This is the measure of the singleton $\{t\}$, which is the atom of $\mu$ at $t$. Note the convention choice: we define $a(t) = \mu((0,t])$ rather than $a(t) = \mu([0,t])$ precisely to ensure $a(0) = 0$. The two conventions differ by the constant $\mu(\{0\})$ and yield the same signed measure on the half-open intervals $(s,t]$.[/guided]

[step:Show $a$ is in $BV[0,T]$ with $V_a(T) \leq |\mu|([0,T])$] We verify that the total variation of $a$ over $[0,T]$ is finite. For any partition $0 = t_0 < t_1 < \cdots < t_n = T$: \begin{align*} \sum_{i=1}^n |a(t_i) - a(t_{i-1})| = \sum_{i=1}^n |\mu((t_{i-1}, t_i])| \leq \sum_{i=1}^n |\mu|((t_{i-1}, t_i]) = |\mu|((0, T]) \leq |\mu|([0,T]). \end{align*} The inequality $|\mu(A)| \leq |\mu|(A)$ holds for every Borel set $A$ by the definition of total variation measure, and the equality $\sum_{i=1}^n |\mu|((t_{i-1}, t_i]) = |\mu|((0, T])$ uses the additivity of the positive measure $|\mu|$ on the disjoint union $(0, T] = \bigsqcup_{i=1}^n (t_{i-1}, t_i]$. Taking the supremum over all partitions: \begin{align*} V_a(T) := \sup_{\text{partitions}} \sum_{i=1}^n |a(t_i) - a(t_{i-1})| \leq |\mu|([0,T]) < \infty. \end{align*} Hence $a \in BV[0,T]$. [/step]

[step:Construct the inverse map: from a cadlag BV function $a$ to a signed measure $\mu$]Let $a: [0,T] \to \mathbb{R}$ be cadlag with $a \in BV[0,T]$ and $a(0) = 0$. Define the total variation function \begin{align*} V_a: [0,T] &\to [0, \infty), \\ t &\mapsto \sup\!\left\{\sum_{i=1}^n |a(t_i) - a(t_{i-1})| : 0 = t_0 < t_1 < \cdots < t_n = t\right\}, \end{align*} and the Jordan-type decomposition: \begin{align*} a^+(t) := \frac{1}{2}(V_a(t) + a(t)), \qquad a^-(t) := \frac{1}{2}(V_a(t) - a(t)). \end{align*} [claim:$a^+$ and $a^-$ are cadlag, non-decreasing, and satisfy $a^+(0) = a^-(0) = 0$] Since $a$ is cadlag and $V_a$ is cadlag (the total variation function of a cadlag BV function inherits right-continuity with left-limits), both $a^+$ and $a^-$ are cadlag as linear combinations of cadlag functions. At $t = 0$: $V_a(0) = 0$ and $a(0) = 0$, so $a^+(0) = a^-(0) = 0$. For monotonicity: for $s < t$, the increment $a^+(t) - a^+(s) = \frac{1}{2}(V_a(t) - V_a(s) + a(t) - a(s))$. Since $V_a(t) - V_a(s) \geq |a(t) - a(s)|$ (the total variation over $[s,t]$ dominates any single increment), we have $V_a(t) - V_a(s) \geq -(a(t) - a(s))$, hence $a^+(t) - a^+(s) \geq 0$. Similarly, $a^-(t) - a^-(s) = \frac{1}{2}(V_a(t) - V_a(s) - (a(t) - a(s))) \geq 0$ since $V_a(t) - V_a(s) \geq a(t) - a(s)$. [/claim] [proof] Contained in the claim statement above. [/proof] Since $a^+$ and $a^-$ are cadlag, non-decreasing functions with $a^+(0) = a^-(0) = 0$, each determines a unique finite positive Borel measure on $[0,T]$ via the Lebesgue-Stieltjes construction: \begin{align*} \mu^+((s,t]) := a^+(t) - a^+(s), \qquad \mu^-((s,t]) := a^-(t) - a^-(s), \end{align*} for $0 \leq s < t \leq T$, extended to all of $\mathcal{B}([0,T])$ by the Caratheodory extension theorem. Define the signed measure $\mu := \mu^+ - \mu^-$.[/step]

[guided]The Lebesgue-Stieltjes construction associates a unique Borel measure to any right-continuous non-decreasing function $F$ with $F(0) = 0$ via $\mu_F((s,t]) = F(t) - F(s)$. The Caratheodory extension theorem extends this from the semiring of half-open intervals to the full Borel $\sigma$-algebra $\mathcal{B}([0,T])$. The extension is unique because $\mu_F$ is $\sigma$-finite (in fact finite, since $\mu_F([0,T]) = F(T) < \infty$). Why do we need $a^+(0) = a^-(0) = 0$? This ensures $\mu^\pm(\{0\}) = a^\pm(0) - a^\pm(0^-) = 0$ (since $a^\pm$ is right-continuous and starts at $0$), which is consistent with the convention $a(0) = 0$. The decomposition $a = a^+ - a^-$ is the Jordan decomposition of the BV function $a$: $a^+$ captures the "upward" variation and $a^-$ the "downward" variation, with $a^+ + a^- = V_a$.[/guided]

[step:Verify the two maps are inverses of each other]**Forward then inverse.** Start with a signed measure $\mu$ on $[0,T]$. The forward map gives $a(t) = \mu((0,t])$. The inverse map decomposes $a = a^+ - a^-$ and builds $\tilde{\mu} = \mu^+ - \mu^-$ with $\tilde{\mu}((s,t]) = a(t) - a(s) = \mu((0,t]) - \mu((0,s]) = \mu((s,t])$. Since $\tilde{\mu}$ and $\mu$ agree on all half-open intervals $(s,t] \subseteq [0,T]$, and these intervals generate $\mathcal{B}([0,T])$ and form a $\pi$-system, the Uniqueness of Extension Theorem (or $\pi$-$\lambda$ theorem applied to the signed measures $\tilde\mu^+, \tilde\mu^-, \mu^+_J, \mu^-_J$ where $\mu = \mu^+_J - \mu^-_J$ is the Jordan decomposition of $\mu$) gives $\tilde{\mu} = \mu$ on all of $\mathcal{B}([0,T])$. **Inverse then forward.** Start with a cadlag BV function $a$ with $a(0) = 0$. The inverse map gives $\mu$ with $\mu((s,t]) = a(t) - a(s)$. The forward map gives $\tilde{a}(t) = \mu((0,t]) = a(t) - a(0) = a(t)$ for $t > 0$, and $\tilde{a}(0) = 0 = a(0)$. Hence $\tilde{a} = a$.[/step]

[guided]The key subtlety in the "forward then inverse" direction is why agreement on half-open intervals suffices to conclude agreement on all Borel sets. The half-open intervals $(s,t]$ with $0 \leq s < t \leq T$ form a $\pi$-system (closed under finite intersections: $(s_1, t_1] \cap (s_2, t_2] = (\max(s_1, s_2), \min(t_1, t_2)]$ when non-empty). They generate $\mathcal{B}([0,T])$. For positive measures, agreement on a $\pi$-system that generates the $\sigma$-algebra implies agreement on the full $\sigma$-algebra, provided both measures are $\sigma$-finite (in fact finite here). For signed measures, we apply this to each pair $(\tilde\mu^+, \mu^+_J)$ and $(\tilde\mu^-, \mu^-_J)$ separately — but we must be careful that the Jordan decompositions of $\tilde\mu$ and $\mu$ may differ. Instead, we argue: $\tilde\mu$ and $\mu$ agree on the $\pi$-system of half-open intervals, so $\tilde\mu^+ + \mu^-_J = \mu^+_J + \tilde\mu^-$ on this $\pi$-system (both sides equal $\tilde\mu^+ + \mu^-_J$ evaluated on $(s,t]$). Since both sides are now positive measures that agree on a generating $\pi$-system, they agree on $\mathcal{B}([0,T])$, giving $\tilde\mu = \mu$.[/guided]

[step:Verify the total variation identity $V_\mu([0,t]) = |\mu|([0,t])$] We show that the total variation of the function $a$ over $[0,t]$ equals $|\mu|((0,t])$, the total variation measure of $\mu$ evaluated on $(0,t]$. For any partition $0 = t_0 < t_1 < \cdots < t_n = t$: \begin{align*} \sum_{i=1}^n |a(t_i) - a(t_{i-1})| = \sum_{i=1}^n |\mu((t_{i-1}, t_i])|. \end{align*} Taking the supremum over all partitions on the left gives $V_a(t)$. On the right, the supremum over all finite partitions of $(0,t]$ into disjoint Borel sets gives $|\mu|((0,t])$ by the definition of total variation measure. **Upper bound.** Since partitions into half-open intervals are a special case of partitions into disjoint Borel sets: \begin{align*} V_a(t) = \sup_{\text{partitions}} \sum_{i=1}^n |\mu((t_{i-1}, t_i])| \leq \sup_{\substack{(0,t] = \bigsqcup_{j=1}^m E_j \\ E_j \in \mathcal{B}}} \sum_{j=1}^m |\mu(E_j)| = |\mu|((0,t]). \end{align*} **Lower bound.** Conversely, for any finite Borel partition $(0,t] = \bigsqcup_{j=1}^m E_j$, we may refine each $E_j$ by intersecting with the half-open intervals of an arbitrarily fine partition. For any $\varepsilon > 0$ and any Borel partition $\{E_j\}$, there exists a partition $0 = t_0 < t_1 < \cdots < t_n = t$ fine enough that: \begin{align*} \sum_{j=1}^m |\mu(E_j)| \leq \sum_{i=1}^n |\mu((t_{i-1}, t_i])| + \varepsilon, \end{align*} by approximating each $E_j$ with unions of half-open intervals (using regularity of finite Borel measures). Taking the supremum over Borel partitions and then $\varepsilon \to 0$ gives $|\mu|((0,t]) \leq V_a(t)$. Combining both bounds: $V_a(t) = |\mu|((0,t])$, which gives $V_\mu([0,t]) = |\mu|([0,t])$ (with the convention that both sides include or exclude $\{0\}$ consistently, noting $a(0) = 0$ means there is no contribution at $t = 0$). This completes the proof of the bijection and the total variation identity. [/step]