Proof of Composition of a Carathéodory–Lipschitz Function with a Measurable Path

Throughout, a function $[a,b]\to\mathbb{R}^{n}$ is measurable if and only if all its components are (Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (i)); a function into $\mathbb{R}^{n}$ has its component functions, and is the only function with them (Component Maps of a Map into $\mathbb{R}^{q}$, Entry Maps of a Map into Matrices, and Maps Assembled from Them (i), Component Maps of a Map into $\mathbb{R}^{q}$, Entry Maps of a Map into Matrices, and Maps Assembled from Them (iii)); and $F_{j}$ denotes the $j$-th component of $F$. By Carathéodory–Lipschitz Right-Hand Sides (iv), $F$ is Carathéodory–Lipschitz with constant $\Lambda$.

Proof of (i). Let $x$ be measurable; its components $x_{1},\dots,x_{n}$ are measurable.

Step 1: paths with finitely many values. For $k\in[n]$ and $m\in\mathbb{N}$ let $t_{m,k}=s_{m}(x_{k}^{+})-s_{m}(x_{k}^{-})$ be the function of Approximation of Measurable Functions by Dyadic Simple Functions (iii) for $x_{k}$: it is a simple function, and

$$t_{m,k}(r)\to x_{k}(r)\quad(m\to\infty)\qquad\text{for every }r\in[a,b].$$

Let $x^{(m)}:[a,b]\to\mathbb{R}^{n}$ be the function with components $t_{m,1},\dots,t_{m,n}$. Each $x^{(m)}$ is measurable, as its components are simple, and its image $V_{m}$ is finite, being contained in $W_{m}^{[n]}$ for the finite union $W_{m}$ of the images of $t_{m,1},\dots,t_{m,n}$ (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (i), Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (i), Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (x), Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vii), Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (ix), Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iv)). By $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (ii), the coordinatewise convergence above gives

$$x^{(m)}(r)\to x(r)\quad\text{in }\mathbb{R}^{n}\qquad\text{for every }r\in[a,b].$$

Step 2: the composed approximations are measurable. Fix $m\in\mathbb{N}$ and $B\in\mathcal{B}(\mathbb{R}^{n})$. Since every $r\in[a,b]$ satisfies $x^{(m)}(r)=c$ for exactly one $c\in V_{m}$,

$$\bigl\{r\in[a,b]:F\bigl(r,x^{(m)}(r)\bigr)\in B\bigr\}=\bigcup_{c\in V_{m}}\Bigl(\bigl\{r\in[a,b]:x^{(m)}(r)=c\bigr\}\cap\bigl\{r\in[a,b]:F(r,c)\in B\bigr\}\Bigr).$$

The first set in each intersection is the preimage under the measurable function $x^{(m)}$ of $\{c\}$, which lies in $\mathcal{B}(\mathbb{R}^{n})$ by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (iii); the second is the preimage of $B$ under $r\mapsto F(r,c)$, which is measurable by Carathéodory–Lipschitz Right-Hand Sides (i). Both therefore lie in $\mathcal{B}_{[a,b]}$ (Measurable Map, and the Set of Measurable Maps (iv), Measurable Map, and the Set of Measurable Maps (i)), and so does the union of their intersections over the finite set $V_{m}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix), and induction over the finite set $V_{m}$ by Induction over Finite Sets, and Predicative Formulas for the Segments [n] and for Finiteness (iii), from Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vi) with Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii)). Hence $F(\cdot,x^{(m)})$ is measurable, and so is each of its components $r\mapsto F_{j}(r,x^{(m)}(r))$, $j\in[n]$.

Step 3: passage to the limit. Let $r\in[a,b]$. By Carathéodory–Lipschitz Right-Hand Sides (ii),

$$\bigl\lVert F\bigl(r,x^{(m)}(r)\bigr)-F\bigl(r,x(r)\bigr)\bigr\rVert\le\Lambda\,\bigl\lVert x^{(m)}(r)-x(r)\bigr\rVert\longrightarrow0\qquad(m\to\infty),$$

so $F(r,x^{(m)}(r))\to F(r,x(r))$ in $\mathbb{R}^{n}$, and by $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (ii) $F_{j}(r,x^{(m)}(r))\to F_{j}(r,x(r))$ for every $j\in[n]$. Fix $j$. The functions $r\mapsto F_{j}(r,x^{(m)}(r))$, $m\in\mathbb{N}$, form a sequence of measurable functions (Step 2) converging at every point of $[a,b]$, so by Suprema, Pointwise Limits and the Convergence Set of Measurable Real-Valued Maps (ii) their pointwise limit is measurable; by uniqueness of limits (The Limit of a Convergent Sequence (i)) that limit is $r\mapsto F_{j}(r,x(r))$. Thus all components of $F(\cdot,x)$ are measurable, and so is $F(\cdot,x)$.

Proof of (ii). Let $x$ be measurable with $\lVert x(r)\rVert\le M$ for every $r$. For $r\in[a,b]$, the triangle inequality (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi)) and Carathéodory–Lipschitz Right-Hand Sides (ii) give

$$\bigl\lVert F\bigl(r,x(r)\bigr)\bigr\rVert\le\lVert F(r,0)\rVert+\bigl\lVert F\bigl(r,x(r)\bigr)-F(r,0)\bigr\rVert\le\lVert F(r,0)\rVert+\Lambda\,\lVert x(r)\rVert\le\lVert F(r,0)\rVert+\Lambda M .$$

The function $r\mapsto\lVert F(r,0)\rVert$ lies in $\mathcal{L}^{1}([a,b])$ by Carathéodory–Lipschitz Right-Hand Sides (iii) and Rules for Integrals of Vector- and Matrix-Valued Functions (viii), and the constant function with value $\Lambda M$, being continuous on $[a,b]$, lies in $\mathcal{L}^{1}([a,b])$ by Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii); so their sum $g$ lies in $\mathcal{L}^{1}([a,b])$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)). By (i) and Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (iii), the function $r\mapsto\lVert F(r,x(r))\rVert$ is measurable, and it is bounded by $g$; hence it lies in $\mathcal{L}^{1}([a,b])$ by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (v), and $F(\cdot,x)\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ by Rules for Integrals of Vector- and Matrix-Valued Functions (i).

Proof of (iii). Let $x$ be continuous on $[a,b]$. Its components are continuous on $[a,b]$ (Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (i), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v)), hence measurable by Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii), so $x$ is measurable. The function $r\mapsto\lVert x(r)\rVert$ is continuous on $[a,b]$ (Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iii), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v)), hence bounded by Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii): there is $M\in\mathbb{R}$ with $\lVert x(r)\rVert\le M$ for every $r\in[a,b]$. Now (ii) applies.

Proof of (iv). For $r\in[a,b]$ this is Carathéodory–Lipschitz Right-Hand Sides (ii) with $x(r)$ and $y(r)$ in place of $y$ and $z$. $\blacksquare$