**Proof plan.** The forward direction (recovering $u$ from $Du$) uses the right-continuous representative and the fact that $Du$ determines $u$ up to a constant. The converse (every finite signed measure is a $BV$ [derivative](/page/Derivative)) is a direct verification using the definition of [distributional derivative](/page/Distributional%20Derivative). The key tool throughout is the identification of the distributional derivative with the Lebesgue–Stieltjes measure. **Step 1: Forward direction — recovery formula.** [claim: Recovery From The Derivative Measure] If $u \in BV(I)$, then $u$ has a right-continuous representative $\tilde{u}$ satisfying $\tilde{u}(x) = \tilde{u}(a^+) + Du((a, x])$. [/claim] [proof] The distributional derivative $Du$ is a finite signed Radon measure on $I$. Define $v(x) := Du((a, x])$ for $x \in I$. Then $v$ is right-continuous (since $Du((a, x_n]) \to Du((a, x])$ when $x_n \downarrow x$, by [continuity](/page/Continuity) of measures from above) and has bounded variation (since $|Dv|(I) = |Du|(I) < \infty$). The distributional derivative of $v$ is $Du$: for any $\phi \in C_c^\infty(I)$, \begin{align*} -\int_I \phi \, dDu = \int_I \phi'(x) v(x) \, d\mathcal{L}^1(x), \end{align*} by [integration by parts](/theorems/210) for Lebesgue–Stieltjes [integrals](/page/Integral) (the [boundary](/page/Boundary) terms vanish since $\phi$ has compact support in $I$). Since $D(u - v) = Du - Dv = 0$ as a distribution, $u - v$ is constant $\mathcal{L}^1$-a.e. Setting $\tilde{u} = v + c$ with $c = \lim_{x \to a^+} u(x) - v(a^+)$ gives the right-continuous representative. [/proof] **Step 2: Converse — every measure is a BV derivative.** [claim: Every Measure Is A Derivative] If $\mu$ is a finite signed Radon measure on $I$, then $v(x) := \mu((a, x])$ belongs to $BV(I)$ with $Dv = \mu$. [/claim] [proof] The [function](/page/Function) $v$ is right-continuous by the same continuity argument as above. For any $\phi \in C_c^\infty(I)$, integration by parts gives: \begin{align*} \int_I v(x) \phi'(x) \, d\mathcal{L}^1(x) = -\int_I \phi(x) \, d\mu(x), \end{align*} which is the definition of $Dv = \mu$ in the [distributional](/page/Distribution) sense. Since $|\mu|(I) < \infty$, the [Dual Characterisation](/theorems/591) gives $v \in BV(I)$ with $|Dv|(I) = |\mu|(I) < \infty$. [/proof] **Step 3: Bijective correspondence.** Combining: the map $u \mapsto Du$ from $BV(I)/\mathbb{R}$ (right-continuous representatives modulo constants) to the space of finite signed Radon measures $\mathcal{M}(I)$ is a bijection, with inverse $\mu \mapsto v(x) = \mu((a, x])$. This is the measure-theoretic version of the classical fundamental theorem.