# Measurable selection for maximum process

Let $$(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P})$$ be a complete filtered probability space, where $$(\mathcal{F}_t)_{t\geq 0}$$ is the completed Brownian filtrate. Suppose that $$\Phi(t,x)$$ is the strong solution for the Stratonovich SDE, say on $$\mathbb{T}^2$$,

$$\Phi(t,x) = \Phi(0,x) + \int_0^t u(\Phi(s,x))ds + \int_0^t \sigma(\Phi(s,x))\circ dW(s), \tag{1}$$

where $$u$$ and $$\sigma$$ are respectively log-Lipschitz and smooth vector fields on $$\mathbb{T}^2$$ and $$W$$ is Brownian motion. Now suppose that $$c(t)$$ is a Hölder continuous stochastic process with state space $$\mathbb{T}^2$$, and consider the stochastic process

$$R(t) := \sup_{x\in \bar{B}(0,r)} |\Phi(t,x)-c(t)|^2. \tag{2}$$

Since $$\Phi$$ is spatially continuous (in fact, it has some Hölder regularity due to $$u$$ being log-Lipschitz) and $$\bar{B}(0,r)$$ is compact, we know that for each $$(t,\omega)$$ fixed, there exists $$x(t,\omega)\in \bar{B}(0,r)$$ such that

$$R(t,\omega) = |\Phi(t,x(t,\omega),\omega)-c(t,\omega)|^2. \tag{3}$$

Evidently, there may be more than one point $$x(t,\omega)$$ satisfying (3); so we have the multi-valued function

$$(t,\omega) \mapsto \{x\in \bar{B}(0;r) : |\Phi(t,x,\omega)-c(t,\omega)|^2 - R(t,\omega)=0\}, \tag{4}$$

where the values of this function are compact.

My question is the following.

Question. Is there a find a selection $$x(t,\omega)$$ in (3) so that the stochastic process $$t \mapsto \Phi(t,x(t))$$ is measurable and adapted?

Write $$K(t,\omega) := \{x\in \bar{B}(0;r) : |\Phi(t,x,\omega)-c(t,\omega)|^2 - R(t,\omega)=0\} \,.$$ Since this set is compact, choosing the lexicographically smallest point $$x^*(t,\omega)$$ in it will yield a measurable selection. For example, if this compact set is in $${\bf R}^2$$, and $$P_i$$ is projection to the $$i$$'th coordinate, let $$x^*_1(t,\omega):=\min P_1(K(t,\omega))$$ and $$x^*_2(t,\omega):=\min P_2(\{x \in K(t,\omega) : P_1(x)=x^*_1(t,\omega)\}) \,.$$

Let me add some detail in response to the comment below. For every real $$\alpha$$, the event $$\{\omega: x^*_1(t,\omega) \le \alpha\}$$ means that for every rational $$\epsilon>0$$ there exists a a rational point $$q=(q_1,q_2)$$ in $$B(0,r)$$ such that $$q_1<\alpha+\epsilon$$ and $$|\Phi(t,q,\omega)-c(t,\omega)|^2 >R(t,\omega)-\epsilon \,.$$ A similar but a bit more involved formula applies to $$\{\omega: x^*_2(t,\omega) \le \beta \}$$.

• Thank you for your comment, Professor Peres, but I am still not quite seeing why it follows that the map $(t,\omega) \mapsto x^*(t,\omega)$ is measurable. It seems like to show this, we want to show that the map $(t,\omega) \mapsto K(t,\omega)$ is measurable where the co-domain is the space $\mathcal{K}(\mathbb{T}^2)$ of non-empty compact subsets of $\mathbb{T}^2$ equipped with the Hausdorff distance and the induced Borel $\sigma$-algebra. However, showing this last point isn't obvious to me. – Matt Rosenzweig Jun 22 '19 at 15:26

This answer is an expansion on the answer of Yuval Peres which addresses the concerns raised in my comment.

We will show the existence of a progressively measurable stochastic process $$t\mapsto x^*(t)\in\mathbb{T}^2$$ satisfying $$|\Phi(t,x^*(t,\omega),\omega)-c(t,\omega)|^2 - R(t,\omega) =0 \tag{1}$$ in several steps.

Let $$\mathcal{K}(\mathbb{T}^2)$$ denote the set of non-empty compact subsets of $$\mathbb{T}^2$$ equipped with the Hausdorff metric $$d_H$$. It is well-known that $$(\mathcal{K}(\mathbb{T}^2), d_H)$$ is a compact metric space. We turn $$\mathcal{K}(\mathbb{T}^2)$$ into a measurable space by endowing it with the Borel $$\sigma$$-algebra $$\mathcal{B}(\mathcal{K}(\mathbb{T}^2))$$.

Lemma 1. The $$\mathcal{K}(\mathbb{T}^2)$$-valued stochastic process $$t \mapsto \bar{B}(c(t), R(t))$$ is progressively measurable.

Proof. Since $$c(\cdot)$$ and $$R(\cdot)$$ are both $$\mathbb{T}^2$$- and $$\mathbb{R}$$-valued progressively measurable processes respectively, it follows that that $$(c(\cdot),R(\cdot))$$ is also progressively measurable. It follows from the triangle inequality that the map $$(\mathbb{T}^2, [0,\infty))\rightarrow (\mathcal{K}(\mathbb{T}^2),d_H), \quad (c,R)\mapsto \partial B(c,R)$$ is continuous, which in turn implies the desired conclusion. $$\Box$$

Although I omitted this detail in my original post, in application the stochastic flow $$\Phi$$ has the following property: for almost every $$\omega$$, the map $$x\mapsto \Phi(t,x,\omega)$$ is a homeomorphism of $$\mathbb{T}^2$$ for all $$t\geq 0$$. Moreover, the $$C(\mathbb{T}^2)$$-valued stochastic process $$t \mapsto \Phi^{-1}(t,\cdot)\in C(\mathbb{T}^2)$$ is progressively measurable. This fact leads us to the next lemma, the proof of which is an easy exercise.

Lemma 2. The map $$\Psi_1: (C(\mathbb{T}^2),\|\cdot\|_\infty) \times (\mathcal{K}(\mathbb{T}^2),d_H) \rightarrow (\mathcal{K}(\mathbb{T}^2),d_H), \qquad (f,A)\mapsto f(A)$$ is continuous.

We claim that the $$\mathcal{K}(\mathbb{T}^2)$$-valued process $$\tilde{K}(t) := \Phi^{-1}(t,\cdot)(\partial B(c(t),R(t))):= \{y\in\mathbb{T}^2: y=\Phi^{-1}(t,x) \text{ for some } x\in \partial B(c(t),R(t))\}$$ is progressively measurable. Indeed, fix $$t\geq 0$$. Evidently, the map $$\Psi_2:[0,t]\times \Omega \rightarrow C(\mathbb{T}^2)\times \mathcal{K}(\mathbb{T}^2), \quad (s,\omega) \mapsto (\Phi^{-1}(s,\cdot,\omega),\partial B(c(s,\omega), R(s,\omega)))$$ is $$\mathcal{B}([0,t]) \otimes \mathcal{B}(\mathcal{K}(\mathbb{T}^2))$$-measurable. So the claim follows from writing $$\tilde{K}(s,\omega) = \Psi_1\circ \Psi_2(s,\omega)$$ and using Lemma 2.

Now $$\tilde{K}(t,\omega)$$ isn't quite the set $$K(t,\omega)$$ from Yuval Peres's answer (it's larger); however, this doesn't matter for the following reason. If $$x\in\tilde{K}(t,\omega)$$, then I claim that $$|x|\geq r$$. Otherwise, since $$\Phi(t,\cdot,\omega)$$ is a homeomorphism, we could find an open ball $$B(x,\delta)\subset B(0,r)$$ such that $$\Phi(t,\cdot,\omega)(B(x,\delta))$$ is open. But by definition of $$\tilde{K}(t,\omega)$$, we have that $$|\Phi(t,x,\omega)-c(t,\omega)|=R(t,\omega)$$, which implies that there exists $$y\in B(x,\delta)$$ such that $$|\Phi(t,y,\omega)-c(t,\omega)|>R(t,\omega).$$ This last inequality contradicts the definition of $$R(t,\omega)$$. Consequently, the lexicographic minimum of $$K(t,\omega)$$, denoted by $$x^*(t,\omega)$$, equals the lexicographic minimum of $$\tilde{K}(t,\omega)$$.

To see that the process $$t\mapsto x^*(t)$$ is progressively measurable, we rely on the following lemma.

Lemma 3. Let $$K$$ be a non-empty compact set. We denote the lexicographic minimum of $$K$$ by $$x_K^*$$. Then the map $$(\mathcal{K}(\mathbb{T}^2), d_H) \rightarrow (\mathbb{T}^2,|\cdot|), \quad K\mapsto x_K^*$$ is continuous.

Proof. Suppose that $$d_H(K_n,K)\rightarrow K$$. Given $$\epsilon>0$$, let $$N\in\mathbb{N}$$ be sufficiently so that $$d_H(K_n,K)\leq \epsilon$$ for all $$n\geq N$$. For each $$n\geq N$$, there exists $$y_{n,K}\in K$$ such that $$|x_{K_n}^*-y_{n,K}| \leq \epsilon,$$ which implies by definition of $$d_H$$ that $$x_{K,j}^* \leq x_{K_n,j}^* + \epsilon, \qquad j=1,2.$$ Similarly, there exists $$z_{K_n}\in K_n$$ such that $$|x_{K}^* - z_{K_n}| \leq \epsilon,$$ which implies that $$x_{K_n,j}^* \leq x_{K,j}^* + \epsilon, \qquad j=1,2.$$ Hence, $$|x_{K_n}^*-x_{K}^*|\leq 2\epsilon$$, and since $$\epsilon>0$$ was arbitrary, the proof of the lemma is complete. $$\Box$$

Applying Lemma 3, we conclude that the stochastic process $$t\mapsto x_{\tilde{K}(t)}^*$$ is progressively measurable.