[proofplan] We fix the spacetime point $(t,x)$ and name its interval of dependence. [D'Alembert's formula](/theorems/665) is then inspected term by term: the displacement part contains only the two endpoint values of $g$, while the velocity part is a [Lebesgue integral](/page/Lebesgue%20Integral) over exactly that interval. Equality of the relevant endpoint values and equality of the velocity data on the interval force equality of the corresponding formula values; setting those data equal to zero gives the vanishing conclusion. [/proofplan] [step:Fix the interval selected by the spacetime point] Fix $t \geq 0$ and $x \in \mathbb{R}$. Define the closed interval $I_{t,x} \subset \mathbb{R}$ by \begin{align*} I_{t,x} = [x - ct, x + ct]. \end{align*} Since $h \in L^1_{\mathrm{loc}}(\mathbb{R})$ and $I_{t,x}$ is compact, the integral \begin{align*} \int_{x - ct}^{x + ct} h(y)\, d\mathcal{L}^1(y) \end{align*} is well-defined. [guided] Fix $t \geq 0$ and $x \in \mathbb{R}$. The spacetime point $(t,x)$ determines exactly one spatial interval, namely \begin{align*} I_{t,x} = [x - ct, x + ct]. \end{align*} This interval is compact because it is a closed bounded interval in $\mathbb{R}$. The hypothesis $h \in L^1_{\mathrm{loc}}(\mathbb{R})$ means that $h$ is integrable over every compact subset of $\mathbb{R}$, so in particular $h$ is integrable over $I_{t,x}$. Therefore the velocity term \begin{align*} \int_{x - ct}^{x + ct} h(y)\, d\mathcal{L}^1(y) \end{align*} is a well-defined Lebesgue integral. [/guided] [/step] [step:Compare two D'Alembert formulas with the same data on the interval] Let $\tilde{g}: \mathbb{R} \to \mathbb{R}$ be a function and let $\tilde{h} \in L^1_{\mathrm{loc}}(\mathbb{R})$. Let $\tilde{u}: [0,\infty) \times \mathbb{R} \to \mathbb{R}$ denote the function defined from $\tilde{g}$ and $\tilde{h}$ by the same D'Alembert formula. Suppose \begin{align*} \tilde{g}(x - ct) = g(x - ct), \end{align*} \begin{align*} \tilde{g}(x + ct) = g(x + ct), \end{align*} and $\tilde{h} = h$ $\mathcal{L}^1$-almost everywhere on $I_{t,x}$. Then equality almost everywhere on the integration interval gives \begin{align*} \int_{x - ct}^{x + ct} \tilde{h}(y)\, d\mathcal{L}^1(y) = \int_{x - ct}^{x + ct} h(y)\, d\mathcal{L}^1(y). \end{align*} Substituting these three equalities into the two D'Alembert formulas yields \begin{align*} \tilde{u}(t,x) = u(t,x). \end{align*} Thus $u(t,x)$ is determined only by $g(x - ct)$, $g(x + ct)$, and the restriction of $h$ to $I_{t,x}$. [/step] [step:Set the endpoint and interval contributions equal to zero] Assume now that \begin{align*} g(x - ct) = g(x + ct) = 0 \end{align*} and that $h = 0$ $\mathcal{L}^1$-almost everywhere on $I_{t,x}$. Since equality almost everywhere preserves the value of the Lebesgue integral, \begin{align*} \int_{x - ct}^{x + ct} h(y)\, d\mathcal{L}^1(y) = 0. \end{align*} D'Alembert's formula then gives \begin{align*} u(t,x) = \frac{1}{2}(0 + 0) + \frac{1}{2c} \cdot 0 = 0. \end{align*} This proves the stated finite propagation property at the point $(t,x)$, and since $(t,x)$ was arbitrary, the theorem follows. [/step]