Proof of Existence and Uniqueness for Ordinary Differential Equations with Measurable Time Dependence

Throughout we use the coordinate bound $|x^i|\le|x|$ from the elementary properties of the Euclidean norm, and the following consequence of linearity of the integral: if $0\le r\le t\le T$ and $g$ is bounded and measurable on $[0,T]$, then $\int_{[0,t]}g=\int_{[0,r]}g+\int_{[r,t]}g$, because $\mathbf{1}_{[0,t]}=\mathbf{1}_{[0,r]}+\mathbf{1}_{[r,t]}-\mathbf{1}_{\{r\}}$ and the integral of $g\,\mathbf{1}_{\{r\}}$ vanishes, the single point $r$ having Lebesgue measure zero. Here $\mathbf{1}_D$ is the function equal to $1$ on $D$ and $0$ elsewhere.

Part (a). Let $y:[0,T]\to\mathbb{R}^n$ be continuous and fix $i$. For a natural number $q$ put $t_k=kT/q$ for $k\in\{0,\dots,q\}$ and define $y^{(q)}:[0,T]\to\mathbb{R}^n$ by $y^{(q)}_s=y_{t_k}$ for $s\in[t_k,t_{k+1})$ and $k\in\{0,\dots,q-1\}$, and $y^{(q)}_T=y_T$.

For every real number $c$,

a finite union of intersections of intervals with sets that belong to the trace Borel $\sigma$-algebra by hypothesis 1. Hence $s\mapsto f^i(s,y^{(q)}_s)$ is measurable, and $|f^i(s,y^{(q)}_s)|\le|f(s,y^{(q)}_s)|\le K$ by hypothesis 2.

The interval $[0,T]$ is nonempty, since $T>0$, and compact by Closed Interval $[a,b]$ is Compact in $\mathbb{R}$, so the continuous map $y$ is uniformly continuous on it. Every $s\in[0,T]$ satisfies $|s-u|\le T/q$ for the point $u\in\{t_0,\dots,t_q\}$ with $y^{(q)}_s=y_u$, so $|y^{(q)}_s-y_s|\le\omega(T/q)$ where $\omega$ is a modulus of uniform continuity of $y$, and therefore $|y^{(q)}_s-y_s|\to0$ as $q\to\infty$, uniformly in $s$. By hypothesis 3 and the coordinate bound,

so $f^i(s,y^{(q)}_s)$ converges to $f^i(s,y_s)$ for every $s\in[0,T]$. All these functions are bounded in absolute value by $K$, so by measurability of bounded pointwise limits, $s\mapsto f^i(s,y_s)$ is measurable; it is bounded by $K$, hence integrable over $[0,t]$ for every $t$.

Part (b), existence. Let $C$ denote the set of continuous maps $[0,T]\to\mathbb{R}^n$ equipped with the supremum metric $d_\infty(y,z)=\sup_{t\in[0,T]}|y_t-z_t|$, which is a complete metric space by completeness of the space of continuous vector-valued functions. Define $x^{(0)}_t=x_0$ for all $t$, and recursively

Assuming inductively that $x^{(j)}\in C$, part (a) shows that all these integrals exist. For $0\le r\le t\le T$ the $i$-th component of $x^{(j+1)}_t-x^{(j+1)}_r$ is $\int_{[r,t]}f^i(s,x^{(j)}_s)\,ds$, so by the norm bound for vector-valued integrals and hypothesis 2,

Hence $x^{(j+1)}\in C$, and by induction every iterate lies in $C$.

We show by induction on $j$ that

For $j=0$ this is the display above with $r=0$. Assuming it for $j$, the norm bound, hypothesis 3 and monotonicity give

the last integral being computed, when $t>0$, as a Riemann integral of a continuous integrand, which agrees with the Lebesgue integral by the compact-interval toolkit, and evaluated by the fundamental theorem of calculus, whose hypotheses hold on $[0,t]$ because the function $s\mapsto K\Lambda^{j}s^{j+2}/(j+2)!$ is continuous on $[0,t]$ and differentiable at every point of $(0,t)$ with derivative the integrand; when $t=0$ both sides vanish.

Consequently $d_\infty(x^{(j+1)},x^{(j)})\le K\Lambda^jT^{j+1}/(j+1)!$. The series $\sum_{j\ge0}K\Lambda^jT^{j+1}/(j+1)!$ converges, its partial sums being bounded by $KT\,e^{\Lambda T}$, so for $j<j'$ the triangle inequality gives

a tail of a convergent series, which tends to $0$ as $j\to\infty$. Thus $(x^{(j)})$ is a Cauchy sequence in $C$ and converges to some $x\in C$; in particular $x$ is continuous.

Fix $t\in[0,T]$ and $i$. By hypothesis 3, the coordinate bound and monotonicity,

Since also $x^{(j+1),i}_t\to x^i_t$, letting $j\to\infty$ in the recursion gives

At $t=0$ all integrals vanish, so the value of $x$ at $0$ is $x_0$.

Part (b), uniqueness. Let $x$ and $\tilde{x}$ be continuous maps satisfying the integral equation with the same $x_0$. The map $u(t)=|x_t-\tilde{x}_t|$ is continuous, and $u(0)=0$. Subtracting the two equations componentwise and applying the norm bound and hypothesis 3,

The function $u$ is continuous on $[0,T]$, which is nonempty, since $T>0$, and compact by Closed Interval $[a,b]$ is Compact in $\mathbb{R}$, hence bounded by Extreme Value Theorem on a Compact Subset of a Metric Space and measurable by claim 4 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions; so Gronwall's lemma for bounded measurable functions, applied with constants $0$ and $\Lambda$, gives $u(t)\le0\cdot e^{\Lambda t}=0$ for every $t$. Since $u\ge0$, we get $u\equiv0$, that is $x=\tilde{x}$.

Part (c). For $0\le r\le t\le T$ the same computation as for the iterates gives $|x_t-x_r|\le\int_{[r,t]}|f(s,x_s)|\,ds\le K(t-r)$, and the case $t\le r$ follows by symmetry. $\blacksquare$