Proof of The Set of Controls with Values in a Closed Bounded Set is Nonempty, Bounded, Convex and Closed

Throughout, $\lambda=\lambda_{[0,T]}$ and $\mathcal{B}=\mathcal{B}_{[0,T]}$ are as in the definition of the Lebesgue space, with $\lambda([0,T])=T$ by the interval toolkit, and the norm of a class does not depend on the representative used to compute it, by The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space. Fix $a_{0}\in\mathcal{A}$, which is possible since $\mathcal{A}$ is nonempty.

A normalized representative. We first record: every $\xi\in\mathcal{U}_{\mathcal{A}}$ has a representative taking all its values in $\mathcal{A}$. Indeed, let $u\in\mathcal{L}^{2}([0,T];\mathbb{R}^{m})$ represent $\xi$ and let $N\in\mathcal{B}$ with $\lambda(N)=0$ be such that $u(t)\in\mathcal{A}$ for $t\in[0,T]\setminus N$. Define $\tilde u(t)=u(t)$ for $t\notin N$ and $\tilde u(t)=a_{0}$ for $t\in N$; each component $\tilde u^{i}$ is measurable: for a set $B$ in the Borel $\sigma$-algebra of $\mathbb{R}$,

$$(\tilde u^{i})^{-1}(B)=\bigl((u^{i})^{-1}(B)\setminus N\bigr)\cup N_{B},$$

where $N_{B}=N$ if $a_{0}^{i}\in B$ and $N_{B}=\emptyset$ otherwise; here $(u^{i})^{-1}(B)\in\mathcal{B}$ because $u^{i}$ is measurable, $N\in\mathcal{B}$, and $\mathcal{B}$ is closed under differences and finite unions by the definition of a $\sigma$-algebra. Moreover $\tilde u=u$ off the null set $N$, so $\tilde u\sim u$, hence $\tilde u$ is square-integrable and $[\tilde u]=\xi$; and $\tilde u(t)\in\mathcal{A}$ for every $t$.

Claim 1. Let $a\in\mathcal{A}$ and let $u$ be the constant map with value $a$. Each component $u^{i}$ is constant, hence a simple function on $([0,T],\mathcal{B})$ and in particular measurable, and $|u(t)|^{2}=|a|^{2}$ for every $t$, so $\int_{[0,T]}|u|^{2}\,d\lambda=|a|^{2}T<\infty$. Thus $u\in\mathcal{L}^{2}([0,T];\mathbb{R}^{m})$, and $u(t)\in\mathcal{A}$ for every $t$, so $[u]\in\mathcal{U}_{\mathcal{A}}$ (take $N=\emptyset$).

Claim 2. Let $\xi\in\mathcal{U}_{\mathcal{A}}$ and let $\tilde u$ be a representative with all values in $\mathcal{A}$. Then $|\tilde u(t)|^{2}\le R^{2}$ for every $t$, so by monotonicity of the integral (claim 1 of Linearity and Monotonicity of the Lebesgue Integral)

$$\lVert\xi\rVert^{2}_{L^{2}}=\int_{[0,T]}|\tilde u|^{2}\,d\lambda\le R^{2}T .$$

Both $\lVert\xi\rVert_{L^{2}}$ and $RT^{1/2}$ are nonnegative and their squares satisfy $\lVert\xi\rVert_{L^{2}}^{2}\le(RT^{1/2})^{2}$, so $\lVert\xi\rVert_{L^{2}}\le RT^{1/2}$: otherwise $RT^{1/2}<\lVert\xi\rVert_{L^{2}}$, and multiplying this inequality by the two nonnegative numbers in turn would give $(RT^{1/2})^{2}<\lVert\xi\rVert^{2}_{L^{2}}$.

Claim 3. Let $\xi,\xi'\in\mathcal{U}_{\mathcal{A}}$ and let $s$ be real with $0\le s\le1$. Choose representatives $\tilde u,\tilde u'$ with all values in $\mathcal{A}$. By the operations of The Lebesgue Space of Square-Integrable Vector-Valued Functions on a Compact Interval, $s\tilde u+(1-s)\tilde u'$ is a representative of $s\xi+(1-s)\xi'$, and it is square-integrable because $L^{2}([0,T];\mathbb{R}^{m})$ is a real vector space with these operations by The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space. By the hypothesis of this claim, $s\tilde u(t)+(1-s)\tilde u'(t)\in\mathcal{A}$ for every $t$. Hence $s\xi+(1-s)\xi'\in\mathcal{U}_{\mathcal{A}}$, which is convexity.

Claim 4. The metric open subsets of $H$ determined by $d_{L^{2}}$ form a topology by Metric Open Sets Form a Topology, and likewise for $\mathbb{R}^{m}$ with the Euclidean distance $d_{E}$. Let $\xi$ belong to the closure of $\mathcal{U}_{\mathcal{A}}$ in $(H,d_{L^{2}})$.

For $n\in\mathbb{N}$ let

$$B_{n}=\bigl\{v\in\mathcal{L}^{2}([0,T];\mathbb{R}^{m})\;:\;v(t)\in\mathcal{A}\text{ for every }t\in[0,T]\text{ and }d_{L^{2}}(\xi,[v])<\iota(n)^{-1}\bigr\},$$

where $\iota$ is the canonical map of $\mathbb{R}$, so that $\iota(n)^{-1}>0$ by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. By Characterization of the Closure in a Metric Space by Open Balls there is an element of $\mathcal{U}_{\mathcal{A}}$ within distance $\iota(n)^{-1}$ of $\xi$, and by the normalized-representative paragraph it has a representative with all values in $\mathcal{A}$; hence $B_{n}\ne\emptyset$. The sets $B_{n}$ are subsets of $\mathcal{L}^{2}([0,T];\mathbb{R}^{m})$, so Axiom of Countable Choice provides a sequence $(u_{n})_{n\in\mathbb{N}}$ with $u_{n}\in B_{n}$ for every $n$.

Then $0\le\lVert[u_{n}]-\xi\rVert_{L^{2}}<\iota(n)^{-1}$, and the real sequence $(\iota(n)^{-1})_{n}$ has limit $0$ by claim 3 of The Archimedean Property of the Real Numbers; comparing with the constant sequence $0$ and using Order Properties of Limits of Real Sequences, the real sequence $\bigl(\lVert[u_{n}]-\xi\rVert_{L^{2}}\bigr)_{n}$ has limit $0$.

Let $u$ be a representative of $\xi$. By claim 2 of Completeness of the Lebesgue Space of Square-Integrable Vector-Valued Functions there are natural numbers $n_{1}<n_{2}<\dots$ and a set $N\in\mathcal{B}$ with $\lambda(N)=0$ such that $\bigl(u_{n_{j}}(t)\bigr)_{j}$ converges to $u(t)$ in $(\mathbb{R}^{m},d_{E})$ for every $t\in[0,T]\setminus N$. Fix such a $t$ and let $\varepsilon>0$ be real; there is $j$ with $d_{E}(u(t),u_{n_{j}}(t))<\varepsilon$, and $u_{n_{j}}(t)\in\mathcal{A}$. By Characterization of the Closure in a Metric Space by Open Balls, $u(t)$ lies in the closure of $\mathcal{A}$ in $(\mathbb{R}^{m},d_{E})$; since $\mathcal{A}$ is closed it equals its closure, so $u(t)\in\mathcal{A}$.

Thus $u(t)\in\mathcal{A}$ for every $t\in[0,T]\setminus N$ with $\lambda(N)=0$, so $\xi=[u]\in\mathcal{U}_{\mathcal{A}}$. Therefore the closure of $\mathcal{U}_{\mathcal{A}}$ is contained in $\mathcal{U}_{\mathcal{A}}$; as it also contains $\mathcal{U}_{\mathcal{A}}$ and is closed, $\mathcal{U}_{\mathcal{A}}$ equals its closure and is closed.