Michael selections and Castaing representations with càdlàg functions

Assume for a contradiction that an strictly increasing sequence $\{t_{n}\}_{n\in\mathbb{N}}\subset[0,T]\setminus\Gamma^{-1}(O)$ exists converging to $t$. Take $y\in O\cap\vec{\Gamma}_{t}$ and $\eta>0$ such that $y+2\eta\mathbb{B}\subset O$. Then $d(y,\Gamma_{t_{n}})\geq\eta$ and there is no sequence $\{y_{n}\}_{n\in\mathbb{N}}$ with $y_{n}\in\Gamma_{t_{n}}$ converging to $y$, a contradiction with the definition of $\vec{\Gamma}$. ∎

Take $t\in(0,T]$ and an open set $O\subset\mathbb{R}^{d}$. Assume that $[t-\delta,t)\subset\Gamma^{-1}(O)$ for some $\delta>0$. For $z\in(t-\delta,t)$ denote by $\mathsf{D}(\Gamma\cap O,z)$ the càdlàg selections $\mathtt{y}\in\mathsf{D}(\Gamma)$ with the further property that $\mathtt{y}(s)\in O$ for $s\in[t-\delta,z)$. Define the set

Claim: $A$ is non-empty and for $z^{*}:=\sup A$ we have $z^{*}=t$. After the claim there exists a càdlàg function $\mathtt{y}\in\mathsf{D}(\Gamma)$ with $\mathtt{y}(z)\in\Gamma_{z}\cap O$ for $z\in[t-\delta,t)$. Then, $\mathtt{y}(t-)\in\vec{\Theta}_{t}$ where $\Theta=\Gamma\cap O$ proving the proposition.

Now we verify the claim. Take $y\in\Gamma_{t-\delta}\cap O$ and $\epsilon>0$ such that $y+\epsilon\mathbb{B}\subset O$. Let $\mathtt{y}\in D(\Gamma)$ be such that $\mathtt{y}(t-\delta)\in\Gamma_{t-\delta}\cap(y+\epsilon\mathbb{B})$. There exists $\eta\in(0,\delta)$ such that $\mathtt{y}(z)\in y+\epsilon\mathbb{B}$ for $z\in[t-\delta,t-\delta+\eta)$ since $\mathtt{y}$ is right continuous. Then, $\mathtt{y}\in D(\Gamma\cap O,t-\delta+\eta)$ showing that $(t-\delta+\eta)\in A$. Thus, $A$ is nonempty. Now assume that $z^{*}<t$, let $\mathtt{y}\in\mathsf{D}(\Gamma\cap O,z^{*})$. By the same argument as before we can find a function $\mathsf{g}\in\mathsf{D}(\Gamma)$ with $\mathsf{g}(z)\in\Gamma_{z}\cap O$ for $z\in[z^{*},z^{*}+\delta^{\prime})$ and some $\delta^{\prime}>0$. Thus, $(z^{*}+\delta^{\prime})\wedge t\in A$ with function $\mathtt{y}1_{[0,z^{*})}+\mathsf{g}1_{[z^{*},T]}$, contradicting the definition of $z^{*}$. Hence $z^{*}=t$ and the proof is complete. ∎

Let $\mathcal{A}$ be a countable family of open sets generating the euclidean topology $\tau_{e}$ of $\mathbb{R}^{d}$. For all $t\in D_{1}$ there exists $A_{t}\in\mathcal{A}$ with $A_{t}\cap\Gamma_{t}\neq\emptyset$ and a strictly increasing sequence $z_{n}\upuparrows t$ such that $\mathop{\rm cl}\nolimits A_{t}\cap\Gamma_{z_{n}}=\emptyset$. We check this claim. By way of contradiction assume that for each $A\in\mathcal{A}$ with $A\cap\Gamma_{t}\neq\emptyset$ it happens that for each strictly increasing sequence $z_{n}\upuparrows t$ there exists $n_{A}\geq 0$ such that for $n\geq n_{A}$ we have $\mathop{\rm cl}\nolimits A\cap\Gamma_{z_{n}}\neq\emptyset$. Let us check that in this case we have $\Gamma_{t}\subset\vec{\Gamma}_{t}$. To this end, take $y\in\Gamma_{t}$. For $\epsilon>0$ let $A_{\epsilon}\in\mathcal{A}$ be such that $y\in A_{\epsilon}$ and $\mathop{\rm cl}\nolimits A_{\epsilon}\subset y+\epsilon\mathbb{B}$. For a sequence $z_{n}\upuparrows t$ there exists $n_{\epsilon}\geq 0$ such that for $n\geq n_{\epsilon}$ we have $\mathop{\rm cl}\nolimits A_{\epsilon}\cap\Gamma_{z_{n}}\neq\emptyset$. Thus, it is possible to construct a sequence $\{y_{n}\}_{n\in\mathbb{N}}$ converging to $y$ with $y_{n}\in\Gamma_{z_{n}}$. Hence $y\in\liminf_{n}\Gamma_{z_{n}}$. It follows that $y\in\vec{\Gamma}_{t}$ since the sequence was arbitrary.

Now assume for a contradiction that $D_{1}$ is uncountable. Then, there exists an infinite subset $D_{0}\subset D_{1}$ and $\tilde{A}\in\mathcal{A}$ such that $A_{t}=\tilde{A}$ for all $t\in D_{0}$. Moreover, there exists a point $t_{0}\in D_{0}$ and a strictly decreasing sequence $\{t_{n}\}_{n\in\mathbb{N}}\subset D_{0}$ converging from the right to $t_{0}$, since $D_{1}$ is uncountable. To verify the latter elementary fact, if there are no right limit points of $D_{1}$ then for each $t\in D_{1}\setminus\{T\}$ there exists $\delta_{t}>0$ such that $(t,t+\delta_{t})\subset[0,T]$ and $(t,t+\delta_{t})\cap D_{1}=\emptyset$. This produces a summable uncountable series of strictly positive numbers which is impossible.

As a consequence, for each $n$ there exists a strictly increasing sequence $\{z_{n,k}\}_{k\in\mathbb{N}}$ converging to $t_{n}$ with $\mathop{\rm cl}\nolimits\tilde{A}\cap\Gamma_{z_{n,k}}=\emptyset$, $\tilde{A}\cap\Gamma_{t_{n}}\neq\emptyset$ and $t_{n+1}<z_{n,k}<t_{n}$. Hence $z_{n,1}\downdownarrows t_{0}$ while $\mathop{\rm cl}\nolimits\tilde{A}\cap\Gamma_{z_{n,1}}=\emptyset$ and $\tilde{A}\cap\Gamma_{t_{0}}\neq\emptyset$, in contradiction to $\Gamma$ being $\tau_{r}$-isc. ∎

Let $\mathcal{A}$ be a countable family of open sets generating the euclidean topology of $\mathbb{R}^{d}$. For all $t$ in the set $\{t\in(0,T]\mid\vec{\Gamma}_{t}\not\subset\Gamma_{t}\}$ there exists $A_{t}\in\mathcal{A}$ with $A_{t}\cap\vec{\Gamma}_{t}\neq\emptyset$ while $\mathop{\rm cl}\nolimits A_{t}\cap\Gamma_{t}=\emptyset$. Assume for a contradiction that the set $\{t\in(0,T]\mid\vec{\Gamma}_{t}\not\subset\Gamma_{t}\}$ is uncountable. Then there exists an infinite subset $D_{0}$ and $\tilde{A}\in\mathcal{A}$ such that for all $t\in D_{0}$ we have $A_{t}=\tilde{A}$ and moreover there exists $t_{0}\in D_{0}$ that is approximated from the left by an increasing sequence $\{t_{n}\}_{n\in\mathbb{N}}\subset D_{0}$. Then, for $n$ large enough we have $\tilde{A}\cap\Gamma_{t_{n}}\neq\emptyset$ since $\tilde{A}\cap\vec{\Gamma}_{t_{0}}\neq\emptyset$. This is a clear contradiction with the properties of $\tilde{A}$ which is equal to $A_{t}$ for $t\in D_{0}$. ∎

For $t\in(0,T)\setminus D_{1}$, take $y_{t}\in\Gamma_{t}\cap\vec{\Gamma}_{t}$. There exists $\delta>0$ such that $(t-\delta,t+\delta)\subset(0,T)$ and $(t-\delta,t+\delta)\subset\Gamma^{-1}(y_{t}+\epsilon\mathbb{B})$ due to the fact that $\Gamma$ is $\tau_{r}$-isc and also by Lemma 1 since $t\notin D_{1}$. Hence, the function $\mathtt{y}^{t}:=y_{t}1_{(t-\delta,t+\delta)}$ is a local continuous selection of $\Gamma+\epsilon\mathbb{B}$.

For $t\in D_{1}\setminus\{0,T\}$ there exists $y^{r}_{t}\in\Gamma_{t}$ since $\Gamma$ has full domain and there exists $\delta_{1}>0$ such that $[t,t+\delta_{1})\subset(0,T)$ and $y^{r}_{t}\in\Gamma(z)+\epsilon\mathbb{B}$ for $z\in[t,t+\delta_{1})$ since $\Gamma$ is $\tau_{r}$-isc. Considering that $\vec{\Gamma}$ has full domain take $y^{l}_{t}\in\vec{\Gamma}_{t}$. There exists $\delta_{2}>0$ such that $(t-\delta_{2},t)\subset\Gamma^{-1}(y^{l}_{t}+\epsilon\mathbb{B})\cap(0,T)$ due to Lemma 1. The function $\mathtt{y}^{t}:=y^{l}_{t}1_{(t-\delta_{2},t)}+y^{r}_{t}1_{[t,t+\delta_{1})}$ is a local càdlàg selection of $\Gamma+\epsilon\mathbb{B}$.

For $t=0$ we can construct by similar arguments a local continuous selection. For $t=T$ a local selection exists that will be continuous or càdlàg according to $T\notin D_{1}$ or $T\in D_{1}$.

The constructed intervals define an open covering (in the relative euclidean topology $\tau_{e}$) of the interval $[0,T]$. There exists a $\tau_{e}$-continuous partition of unity subordinated to a locally finite subcovering from which a global càdlàg selection is generated with the required property of continuity outside $D_{1}$. ∎

Take $t\notin D_{1}$ with $t>0$, and $y\in\phi_{t}$. Let $\epsilon^{\prime}\in(0,\epsilon)$ be such that $\left|y-\mathsf{g}(t)\right|<\epsilon^{\prime}$. For $\rho:=(\epsilon-\epsilon^{\prime})$ let $\delta_{1}>0$ be such that $t-\delta_{1}>0$ and for $z\in[t-\delta_{1},t)$ we have $\left|\mathsf{g}(z)-\mathsf{g}(t)\right|<\rho$. There exists $\delta_{2}\in(0,\delta_{1})$ such that $[t-\delta_{2},t)\subset\Gamma^{-1}(y+\eta\mathbb{B})$ due to Lemma 1, since $y\in\vec{\Gamma}_{t}$.

For $z\in[t-\delta_{2},t)$ let $r_{1}=y-\mathsf{g}(t)\in\epsilon^{\prime}\mathbb{B}$ and $r_{2}=\mathsf{g}(t)-\mathsf{g}(z)\in\rho\mathbb{B}$. Then $y-r_{2}=\mathsf{g}(z)+r_{1}$ showing that

which completes the proof. ∎

Let $\epsilon_{0}=1$ and $\epsilon_{i}:=\frac{1}{2^{i}}\epsilon_{i-1}$. By Lemma 7, there exists a càdlàg selection $\mathtt{y}_{1}$ of $\Gamma+\epsilon_{1}\mathbb{B}$ continuous outside $D_{1}$. Assume $\mathtt{y}_{1},\ldots,\mathtt{y}_{k}$ have been constructed with the following properties:

Now we construct a function satisfying (a)-(c) for $i=k+1$. This will produce a sequence $\{\mathtt{y}_{i}\}_{i\in\mathbb{N}}$ converging under the uniform norm. Hence, it converges to a càdlàg function $\mathtt{y}\in\mathsf{D}$ which is continuous outside $D_{1}$. The function $\mathtt{y}$ is then a selection of $\mathop{\rm cl}\nolimits\Gamma$.

Let $\Gamma^{k+1}:=\Gamma\cap(\mathtt{y}_{k}+\epsilon_{k}\mathbb{B})$. The mapping $\Gamma^{k+1}$ is $\tau_{r}$-isc since it is the intersection of $\tau_{r}$-isc mappings; see Lemma 16. It is clearly convex valued. Moreover, $\mathop{\rm dom}\nolimits\Gamma^{k+1}=[0,T]$ since $\mathtt{y}_{k}$ is a selection of $\Gamma+\epsilon_{k}\mathbb{B}$.

Take $t\in(0,T)\setminus D_{1}$ and $y_{t}\in\Gamma_{t}^{k+1}$. Let $\eta:=\frac{1}{2}\epsilon_{k+1}$. There exists $\delta_{1}>0$ such that $t-\delta_{1}>0$ and $y_{t}\in(\Gamma_{z}+\eta\mathbb{B})\cap(\mathtt{y}_{k}(z)+\epsilon_{k}\mathbb{B})$ for $z\in(t-\delta_{1},t)$, due to Lemma 8. There exists $\delta_{2}>0$ such that $t+\delta_{2}<T$ and $(y_{t}+\eta\mathbb{B})\cap\Gamma_{z}^{k+1}\neq\emptyset$ for $z\in[t,t+\delta_{2})$, since $\Gamma^{k+1}$ is $\tau_{r}$-isc. Hence, for $I_{t}:=(t-\delta_{1},t+\delta_{2})$ we have $I_{t}\subset(0,T)$ and for $z\in I_{t}$ there exist $\gamma_{z}\in\Gamma_{z}$, $r_{1},r_{3}\in\eta\mathbb{B}$, $r_{2}\in\epsilon_{k}$ such that

Thus, $y_{t}\in(\Gamma_{z}+\epsilon_{k+1}\mathbb{B})\cap(\mathtt{y}_{k}(z)+2\epsilon_{k}\mathbb{B})$. As a consequence, the function $\mathtt{y}^{t}_{k+1}:=y_{t}1_{I_{t}}$ is a local continuous selection of $\Gamma+\epsilon_{k+1}\mathbb{B}$ and $\mathtt{y}_{k}+2\epsilon_{k}\mathbb{B}$.

Now take $t\in D_{1}\cap(0,T)$. Let $\eta:=\frac{1}{3}\epsilon_{k+1}$ and $O_{1}:=\mathtt{y}_{k}(t-)+(\epsilon_{k}+\eta)\mathbb{B}$. There exists $\delta_{1}>0$ such that $t-\delta_{1}>0$ and $(t-\delta_{1},t)\subset\Gamma^{-1}(O_{1})$ since $\mathtt{y}_{k}$ is a selection of $\Gamma+\epsilon_{k}\mathbb{B}$. Hence, there exists $y^{l}_{t}\in\overrightarrow{\Gamma_{t}\cap O_{1}}$ since $\Gamma$ satisfies \threfass. There exists $\delta_{2}<\delta_{1}$ such that for $z\in(t-\delta_{2},t)$ we have

due to Lemma 1 and we can select $\delta_{2}$ so that $|\mathtt{y}_{k}(t-)-\mathtt{y}_{k}(z)|<\eta$ for $z\in(t-\delta_{2},t)$. Thus, $y^{l}_{t}\in(\Gamma_{z}+\epsilon_{k+1}\mathbb{B})\cap(\mathtt{y}_{k}(z)+2\epsilon_{k}\mathbb{B})$ for $z\in(t-\delta_{2},t)$. Take $y_{t}^{r}\in\Gamma^{k+1}_{t}$. There exists $\delta_{3}>0$ such that $t+\delta_{3}\leq T$ and $[t,t+\delta_{3})\subset(\Gamma^{k+1})^{-1}(y_{t}^{r}+\eta\mathbb{B})$ due to the fact that $\Gamma^{k+1}$ is $\tau_{r}$-isc. Let $I_{t}:=(t-\delta_{2},t+\delta_{3})$ and define a function $\mathtt{y}^{t}_{k+1}$ on $I_{t}$ by $\mathtt{y}^{t}_{k+1}:=y^{l}_{t}1_{(t-\delta_{2},t)}+y^{r}_{t}1_{[t,t+\delta_{3})}$. It is clear that $\mathtt{y}^{t}_{k+1}$ is a local càdlàg selection of $\Gamma^{k+1}+\epsilon_{k+1}\mathbb{B}$ and $\mathtt{y}_{k}+2\epsilon_{k}\mathbb{B}$. For $t\in\{0,T\}$ we can construct local selections by similar arguments.

The constructed intervals $\{I_{t}\}_{t\in[0,T]}$ define an open covering (in the relative euclidean topology $\tau_{e}$) of the interval $[0,T]$. There exists a $\tau_{e}$-continuous partition of unity subordinated to a locally finite subcovering from which a global selection $\mathtt{y}_{k+1}$ can be produced by pasting together the local selections. ∎

If $T\in D_{1}$ we take a selection $\mathtt{y}\in\mathsf{D}(\Gamma)$ continuous outside $D_{1}$ from which we can easily construct a sequence $\{\mathtt{y}_{j}\}_{j\in\mathbb{N}}\subset\mathsf{D}(\Gamma)$ by modifying $\mathtt{y}$ at $T$ such that $\{\mathtt{y}_{j}(T)\}_{j\in\mathbb{N}}$ is dense in $\Gamma_{T}$. This settles down the representation for $t=T$ in case $T\in D_{1}$.

Let $\mathcal{D}=\{y_{m}\}_{m\in\mathbb{N}}$ be a countable dense subset of $\mathbb{R}^{d}$. Take $\epsilon=\frac{1}{2^{k}}$ for $k\in\mathbb{N}$ and $y=y_{m}$ for $y_{m}\in\mathcal{D}$. Assume that $U:=\Gamma^{-1}(y+\epsilon\mathbb{B})$ is non empty. The set $U$ is open in $\tau_{r}$ (relativized to $[0,T]$) and can be expressed as a countable union of intervals of the form $[a,b]$. Indeed, $\tau_{r}$ is hereditary Lindelöf and $U$ can be written as the countable union of intervals $[x,y)\cap[0,T]$ and each of these intervals can themselves be written as the countable union of intervals $[a,b]$. If $T\in U$ we distinguish two cases. If $T\notin D_{1}$ then $T$ can be taken as an element of an interval $[a,T]\subset U$ due to Lemma 1. In this case, we will consider the representation (i) $U=\bigcup_{n\in\mathbb{N}}[a_{n},b_{n}]$ with $a_{n}<b_{n}$, thus, no interval collapses to a point. In the second case $T\in U$ and $T\in D_{1}$ and we consider the representation (ii) $U=\{T\}\cup\bigcup_{n\in\mathbb{N}}[a_{n},b_{n}]$ with $a_{n}<b_{n}<T$ and $b_{n}\notin D_{1}$. Hence, on both cases we do not consider trivial intervals and $b_{n}\notin D_{1}$.

Now take an interval $[a,b]=[a_{n},b_{n}]\subset U$ with $a_{n}<b_{n}$ and $b_{n}\notin D_{1}$. We fix the notation $[a,b]^{c}:=[0,T]\setminus[a,b]$. Let $\phi$ be the mapping defined by

It is simple to verify that the mapping $\phi$ is convex valued, has full domain, and that it is $\tau_{r}$-isc.

We verify \threfass and that $\vec{\phi}$ has full domain. Let $\delta>0$ and $t\in(0,T]$ be such that $(t-\delta,t)\subset\phi^{-1}(O)$ for $O$ an open subset of $\mathbb{R}^{d}$. For $t\notin(a,b]$ there exists $\delta^{\prime}\in(0,\delta)$ such that $(t-\delta^{\prime},t)\subset[a,b]^{c}$ so $\phi=\Gamma$ in this interval and we have $\phi\cap O=\Gamma\cap O$, hence \threfass is clearly satisfied in $(a,b]$. Take now $t\in(a,b]$. There exists $\delta^{\prime\prime}\in(0,\delta)$ such that $(t-\delta^{\prime\prime},t)\subset(a,b]\cap(t-\delta,t)$ and for $z\in(t-\delta^{\prime\prime},t)$ we have that $\phi_{z}\cap O=\Gamma_{z}\cap(y+\epsilon\mathbb{B})\cap O\neq\emptyset$ so by \threfass we have $\overrightarrow{\Gamma\cap(y+\epsilon\mathbb{B})\cap O}=\overrightarrow{\phi\cap O}$ is non empty at $t$. Thus, $\phi$ satisfies \threfass and $\mathop{\rm dom}\nolimits\vec{\phi}=[0,T]$.

The set $\{t\mid\phi_{t}\not\subset\vec{\phi}_{t}\}$ is included in $D_{1}$. Indeed, for $t\notin(a,b]$ there exists $\delta>0$ such that $(t-\delta,t)\subset[a,b]^{c}$ so $\vec{\phi}_{t}=\vec{\Gamma}_{t}$ and $\phi_{t}\subset\Gamma_{t}$. Hence, $\phi_{t}\subset\vec{\phi}_{t}$ whenever $t\notin D_{1}$. Take now $t\in(a,b]\setminus D_{1}$ and $w\in\phi_{t}$. Let $\eta>0$ be such that $w+\eta\mathbb{B}\subset y+\epsilon\mathbb{B}$. Note that $w\in\vec{\Gamma}_{t}$ since $t\notin D_{1}$ and then $w\in\Gamma_{t}\subset\vec{\Gamma}_{t}$. There exists $\delta_{\eta}>0$ such that for $z\in(t-\delta_{\eta},t)$ we have $\Gamma_{z}\cap(w+\eta\mathbb{B})\neq\emptyset$ due to Lemma 1, since $w\in\vec{\Gamma}_{t}$. Hence, $\overrightarrow{\Gamma\cap(w+\eta\mathbb{B})}\neq\emptyset$ at $t$ by \threfass. Given that $\eta$ was arbitrary, we deduce that $w\in\vec{\phi}_{t}$. Hence $\phi_{t}\subset\vec{\phi}_{t}$. This proves the claim.

Hence, there exists a càdlàg selection of $\mathop{\rm cl}\nolimits\phi$ continuous outside $D_{1}$, due to Theorem 9. From this one derives the existence of a selection $\mathtt{y}_{n,\epsilon,y}\in\mathsf{D}(\Gamma)$ with $\mathtt{y}_{n,\epsilon,y}(t)\in y+2\epsilon\mathbb{B}$ for $t\in[a_{n},b_{n}]$, and the continuity in $[0,T]\setminus D_{1}$.

If $T\notin D_{1}$ it is easy to verify that the required representation holds with the countable family $\{\mathtt{y}_{n,2^{-k},y_{m}}\}_{n,k,m}\subset\mathsf{D}(\Gamma)$. Analogously, if $T\in D_{1}$, consider the family $\{\mathtt{y}_{j}\}_{j\in\mathbb{N}}\cup\{\mathtt{y}_{n,2^{-k},y_{m}}\}_{n,k,m}\subset\mathsf{D}(\Gamma)$. ∎