[proofplan] We prove both directions by applying the [Fundamental Theorem of Calculus](/theorems/632) to the continuous vector field obtained by composing $F$ with the curve $x$. In the forward direction, the differential equation gives the derivative of each component of $x$, and integration from $t_0$ to $t$ gives the integral identity, with the oriented convention handling $t<t_0$. In the reverse direction, evaluating the identity at $t_0$ gives the initial condition, and subtracting the integral identities at nearby times gives local integral formulas whose derivatives are recovered by the same theorem. The openness of $U$ is part of the [ordinary differential equation](/page/Ordinary%20Differential%20Equation) setting; the equivalence itself only uses that $x(s)\in U$ and that $F(s,x(s))$ is defined and continuous along the curve. [/proofplan]

[step:Use the oriented integral convention on intervals between $t_0$ and $t$] For $t\in J$, define the compact interval $K_t\subset J$ by \begin{align*} K_t=[\min\{t_0,t\},\max\{t_0,t\}]. \end{align*} The oriented integral convention is \begin{align*} \int_{t_0}^{t} h(s)\,d\mathcal{L}^1(s)=\int_{K_t} h(s)\,d\mathcal{L}^1(s) \end{align*} when $t\ge t_0$, and \begin{align*} \int_{t_0}^{t} h(s)\,d\mathcal{L}^1(s)=-\int_{K_t} h(s)\,d\mathcal{L}^1(s) \end{align*} when $t<t_0$, for every continuous map $h:K_t\to \mathbb{R}^n$. For each fixed $t\in J$, define the map $g_t:K_t\to \mathbb{R}^n$ by \begin{align*} g_t(s)=F(s,x(s)). \end{align*} Since $x:J\to U$ is continuous and $F:I\times U\to \mathbb{R}^n$ is continuous, the map $g_t$ is continuous on $K_t$. [/step]

[step:Integrate the differential equation to obtain the integral identity]Assume first that $x$ is a solution of the initial value problem. Thus $x(t_0)=x_0$, and for every interior point $s$ of $J$, \begin{align*} \dot{x}(s)=F(s,x(s)). \end{align*} Fix $t\in J$. If $t=t_0$, the desired identity reduces to $x(t_0)=x_0$. Suppose $t>t_0$. Write $x=(x_1,\dots,x_n)$ and $F=(F_1,\dots,F_n)$, where each $F_i:I\times U\to\mathbb{R}$ is the $i$th coordinate function of $F$. For each component index $i\in\{1,\dots,n\}$, the scalar function $x_i:K_t\to \mathbb{R}$ is continuous on $K_t$, differentiable on the interior of $K_t$, and satisfies \begin{align*} \dot{x}_i(s)=F_i(s,x(s)) \end{align*} for every interior point $s$ of $K_t$. Since $s\mapsto F_i(s,x(s))$ is continuous on $K_t$, the [Fundamental Theorem of Calculus](/theorems/632), applied to the continuous integrand $s\mapsto F_i(s,x(s))$ and the differentiable function $x_i$, gives \begin{align*} x_i(t)-x_i(t_0)=\int_{t_0}^{t}F_i(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Combining the component identities gives \begin{align*} x(t)=x_0+\int_{t_0}^{t}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} If $t<t_0$, the same componentwise argument on $K_t=[t,t_0]$ gives \begin{align*} x_i(t_0)-x_i(t)=\int_{t}^{t_0}F_i(s,x(s))\,d\mathcal{L}^1(s). \end{align*} By the oriented integral convention, this is equivalent to \begin{align*} x_i(t)-x_i(t_0)=\int_{t_0}^{t}F_i(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Again combining components yields the stated vector identity.[/step]

[guided]Assume that $x$ is a solution of the initial value problem. This means two things: first, \begin{align*} x(t_0)=x_0, \end{align*} and second, at every interior point $s$ of $J$, \begin{align*} \dot{x}(s)=F(s,x(s)). \end{align*} The goal is to recover $x(t)$ by accumulating its velocity from $t_0$ to $t$. Fix $t\in J$. If $t=t_0$, then the integral over the degenerate interval is zero, so the desired identity is exactly the initial condition \begin{align*} x(t_0)=x_0. \end{align*} Now suppose $t>t_0$. Write $x=(x_1,\dots,x_n)$ and $F=(F_1,\dots,F_n)$, where $F_i:I\times U\to\mathbb{R}$ denotes the $i$th coordinate function of $F$. For each component index $i\in\{1,\dots,n\}$, consider the scalar component $x_i:K_t\to\mathbb{R}$. The function $x_i$ is continuous on $K_t=[t_0,t]$, differentiable at every interior point of $K_t$, and its derivative is \begin{align*} \dot{x}_i(s)=F_i(s,x(s)). \end{align*} The right-hand side is continuous in $s$, because it is the $i$th component of the continuous map $s\mapsto F(s,x(s))$. Therefore the [Fundamental Theorem of Calculus](/theorems/632) applies to $x_i$ on $[t_0,t]$: the function $x_i$ is continuous on $[t_0,t]$, differentiable on $(t_0,t)$, and has continuous derivative $s\mapsto F_i(s,x(s))$. It gives \begin{align*} x_i(t)-x_i(t_0)=\int_{t_0}^{t}F_i(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Since this holds for every component $i$, the vector identity follows: \begin{align*} x(t)-x(t_0)=\int_{t_0}^{t}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Using $x(t_0)=x_0$, we obtain \begin{align*} x(t)=x_0+\int_{t_0}^{t}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Finally suppose $t<t_0$. The fundamental theorem of calculus is applied on the ordinary compact interval $[t,t_0]$. For each component $i$ it gives \begin{align*} x_i(t_0)-x_i(t)=\int_{t}^{t_0}F_i(s,x(s))\,d\mathcal{L}^1(s). \end{align*} The oriented integral convention says that reversing the endpoints changes the sign: \begin{align*} \int_{t_0}^{t}F_i(s,x(s))\,d\mathcal{L}^1(s)=-\int_{t}^{t_0}F_i(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Hence \begin{align*} x_i(t)-x_i(t_0)=\int_{t_0}^{t}F_i(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Combining the component identities and using $x(t_0)=x_0$ gives the required integral formula for $t<t_0$ as well.[/guided]

[step:Evaluate the integral identity at the initial time] Conversely, assume that for every $t\in J$, \begin{align*} x(t)=x_0+\int_{t_0}^{t}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Taking $t=t_0$, the oriented integral over the degenerate interval is zero, so \begin{align*} x(t_0)=x_0. \end{align*} Thus the initial condition holds. [/step]

[step:Differentiate the local integral identity at interior points]Let $\tau$ be an interior point of $J$. Choose $\rho>0$ such that \begin{align*} (\tau-\rho,\tau+\rho)\subset J. \end{align*} Define the map $g:(\tau-\rho,\tau+\rho)\to\mathbb{R}^n$ by \begin{align*} g(s)=F(s,x(s)). \end{align*} The map $g$ is continuous because it is the composition of the continuous map $s\mapsto (s,x(s))$ with $F$. For every $t\in(\tau-\rho,\tau+\rho)$, subtracting the assumed integral identity at $\tau$ from the assumed integral identity at $t$ gives \begin{align*} x(t)-x(\tau)=\int_{t_0}^{t}F(s,x(s))\,d\mathcal{L}^1(s)-\int_{t_0}^{\tau}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} By additivity of the oriented integral over intervals, this becomes the local identity \begin{align*} x(t)-x(\tau)=\int_{\tau}^{t}g(s)\,d\mathcal{L}^1(s). \end{align*} Define the local primitive $G:(\tau-\rho,\tau+\rho)\to\mathbb{R}^n$ by \begin{align*} G(t)=\int_{\tau}^{t}g(s)\,d\mathcal{L}^1(s). \end{align*} The [Fundamental Theorem of Calculus](/theorems/632), applied componentwise to the [continuous function](/page/Continuous%20Function) $g$ on compact subintervals of $(\tau-\rho,\tau+\rho)$, gives that $G$ is differentiable at $\tau$ and \begin{align*} \dot{G}(\tau)=g(\tau). \end{align*} Since the local identity gives $x(t)-x(\tau)=G(t)$ for $t$ near $\tau$, the curve $x$ is differentiable at $\tau$ and \begin{align*} \dot{x}(\tau)=\dot{G}(\tau)=g(\tau)=F(\tau,x(\tau)). \end{align*} Thus the differential equation holds at every interior point of $J$.[/step]

[guided]We now prove the reverse implication at an interior point. Let $\tau$ be an interior point of $J$. Choose $\rho>0$ such that \begin{align*} (\tau-\rho,\tau+\rho)\subset J. \end{align*} Define $g:(\tau-\rho,\tau+\rho)\to\mathbb{R}^n$ by \begin{align*} g(s)=F(s,x(s)). \end{align*} This map is continuous because $s\mapsto(s,x(s))$ is continuous from $(\tau-\rho,\tau+\rho)$ to $I\times U$, and $F:I\times U\to\mathbb{R}^n$ is continuous. For every $t\in(\tau-\rho,\tau+\rho)$, subtract the assumed integral identity at $\tau$ from the assumed integral identity at $t$. This gives \begin{align*} x(t)-x(\tau)=\int_{t_0}^{t}F(s,x(s))\,d\mathcal{L}^1(s)-\int_{t_0}^{\tau}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} By additivity of oriented integrals over intervals, the right-hand side is the local integral from $\tau$ to $t$: \begin{align*} x(t)-x(\tau)=\int_{\tau}^{t}g(s)\,d\mathcal{L}^1(s). \end{align*} The point of subtracting at $\tau$ is that the resulting integral only involves values of $g$ on the neighbourhood where $g$ was defined. Define $G:(\tau-\rho,\tau+\rho)\to\mathbb{R}^n$ by \begin{align*} G(t)=\int_{\tau}^{t}g(s)\,d\mathcal{L}^1(s). \end{align*} The [Fundamental Theorem of Calculus](/theorems/632) applies componentwise to $g$: each coordinate function of $g$ is continuous on compact subintervals of $(\tau-\rho,\tau+\rho)$. Hence $G$ is differentiable at $\tau$ and \begin{align*} \dot{G}(\tau)=g(\tau). \end{align*} Because $x(t)-x(\tau)=G(t)$ for $t$ near $\tau$, differentiating at $\tau$ gives \begin{align*} \dot{x}(\tau)=\dot{G}(\tau)=g(\tau)=F(\tau,x(\tau)). \end{align*} Thus the differential equation holds at the arbitrary interior point $\tau$.[/guided]

[step:Take one-sided derivatives at endpoints of $J$] Let $a\in J$ be a left endpoint of $J$ belonging to $J$ and suppose $a$ is an accumulation point of $J$ from the right. For $h>0$ with $a+h\in J$, subtracting the assumed integral identity at $a$ from the assumed integral identity at $a+h$ gives \begin{align*} \frac{x(a+h)-x(a)}{h}=\frac{1}{h}\int_a^{a+h}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} The map $s\mapsto F(s,x(s))$ is continuous at $a$. Hence the one-sided form of the [Fundamental Theorem of Calculus](/theorems/632), applied componentwise to the continuous integrand $s\mapsto F(s,x(s))$, gives \begin{align*} \lim_{h\downarrow 0}\frac{x(a+h)-x(a)}{h}=F(a,x(a)). \end{align*} So the right derivative at the left endpoint exists and satisfies the differential equation. Similarly, if $b\in J$ is a right endpoint of $J$ belonging to $J$ and $b$ is an accumulation point of $J$ from the left, then for $h>0$ with $b-h\in J$, subtracting the assumed integral identity at $b-h$ from the assumed integral identity at $b$ gives \begin{align*} \frac{x(b)-x(b-h)}{h}=\frac{1}{h}\int_{b-h}^{b}F(s,x(s))\,d\mathcal{L}^1(s). \end{align*} Continuity of $s\mapsto F(s,x(s))$ at $b$ and the one-sided form of the [Fundamental Theorem of Calculus](/theorems/632) give \begin{align*} \lim_{h\downarrow 0}\frac{x(b)-x(b-h)}{h}=F(b,x(b)). \end{align*} Equivalently, the left derivative of $x$ at $b$ exists and equals $F(b,x(b))$. Therefore $x$ satisfies the differential equation at endpoints in the required one-sided sense. Together with $x(t_0)=x_0$, this proves that $x$ is a solution of the initial value problem. [/step]