# Jointly measurable and progressively measurable stochastic processes

Jordan Bell
June 18, 2015

## 1 Jointly measurable stochastic processes

Let $E=\mathbb{R}^{d}$ with Borel $\mathscr{E}$, let $I=\mathbb{R}_{\geq 0}$, which is a topological space with the subspace topology inherited from $\mathbb{R}$, and let $(\Omega,\mathscr{F},P)$ be a probability space. For a stochastic process $(X_{t})_{t\in I}$ with state space $E$, we say that $X$ is jointly measurable if the map $(t,\omega)\mapsto X_{t}(\omega)$ is measurable $\mathscr{B}_{I}\otimes\mathscr{F}\to\mathscr{E}$.

For $\omega\in\Omega$, the path $t\mapsto X_{t}(\omega)$ is called left-continuous if for each $t\in I$,

 $X_{s}(\omega)\to X_{t}(\omega),\qquad s\uparrow t.$

We prove that if the paths of a stochastic process are left-continuous then the stochastic process is jointly measurable.11 1 cf. Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 153, Lemma 4.51.

###### Theorem 1.

If $X$ is a stochastic process with state space $E$ and all the paths of $X$ are left-continuous, then $X$ is jointly measurable.

###### Proof.

For $n\geq 1$, $t\in I$, and $\omega\in\Omega$, let

 $X^{n}_{t}(\omega)=\sum_{k=0}^{\infty}1_{[k2^{-n},(k+1)2^{-n})}(t)X_{k2^{-n}}(% \omega).$

Each $X^{n}$ is measurable $\mathscr{B}_{I}\otimes\mathscr{F}\to\mathscr{E}$: for $B\in\mathscr{E}$,

 $\{(t,\omega)\in I\times\Omega:X^{n}_{t}(\omega)\in B\}=\bigcup_{k=0}^{\infty}[% k2^{-n},(k+1)2^{-n})\times\{X_{k2^{-n}}\in B\}.$

Let $t\in I$. For each $n$ there is a unique $k_{n}$ for which $t\in[k_{n}2^{-n},(k_{n}+1)2^{-n})$, and thus $X_{t}^{n}(\omega)=X_{k_{n}2^{-n}}(\omega)$. Furthermore, $k_{n}2^{-n}\uparrow t$, and because $s\mapsto X_{s}(\omega)$ is left-continuous, $X_{k_{n}2^{-n}}(\omega)\to X_{t}(\omega)$. That is, $X^{n}\to X$ pointwise on $I\times\Omega$, and because each $X_{n}$ is measurable $\mathscr{B}_{I}\otimes\mathscr{F}\to\mathscr{E}$ this implies that $X$ is measurable $\mathscr{B}_{I}\otimes\mathscr{F}\to\mathscr{E}$.22 2 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 142, Lemma 4.29. Namely, the stochastic process $(X_{t})_{t\in I}$ is jointly measurable, proving the claim. ∎

Let $\mathscr{F}_{I}=(\mathscr{F}_{t})_{t\in I}$ be a filtration of $\mathscr{F}$. A stochastic process $X$ is said to be adapted to the filtration $\mathscr{F}_{I}$ if for each $t\in I$ the map

 $\omega\mapsto X_{t}(\omega),\qquad\Omega\to E,$

is measurable $\mathscr{F}_{t}\to\mathscr{E}$, in other words, for each $t\in I$,

 $\sigma(X_{t})\subset\mathscr{F}_{t}.$

For a stochastic process $(X_{t})_{t\in I}$, the natural filtration of $X$ is

 $\sigma(X_{s}:s\leq t).$

It is immediate that this is a filtration and that $X$ is adapted to it.

## 3 Progressively measurable stochastic processes

Let $\mathscr{F}_{I}=(\mathscr{F}_{t})_{t\in I}$ be a filtration of $\mathscr{F}$. A function $X:I\times\Omega\to E$ is called progressively measurable with respect to the filtration $\mathscr{F}_{I}$ if for each $t\in I$, the map

 $(s,\omega)\mapsto X(s,\omega),\qquad[0,t]\times\Omega\to E,$

is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$. We denote by $\mathscr{M}^{0}(\mathscr{F}_{I})$ the set of functions $I\times\Omega\to E$ that are progressively measurable with respect to the filtration $\mathscr{F}_{I}$. We shall speak about a stochastic process $(X_{t})_{t\in I}$ being progressively measurable, by which we mean that the map $(t,\omega)\mapsto X_{t}(\omega)$ is progressively measurable.

We denote by $\mathrm{Prog}(\mathscr{F}_{I})$ the collection of those subsets $A$ of $I\times\Omega$ such that for each $t\in I$,

 $([0,t]\times\Omega)\cap A\in\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}.$

We prove in the following that this is a $\sigma$-subalgebra of $\mathscr{B}_{I}\otimes\mathscr{F}$ and that it is the coarsest $\sigma$-algebra with which all progressively measurable functions are measurable.

###### Theorem 2.

Let $\mathscr{F}_{I}=(\mathscr{F}_{t})_{t\in I}$ be a filtration of $\mathscr{F}$.

1. 1.

$\mathrm{Prog}(\mathscr{F}_{I})$ is a $\sigma$-subalgebra of $\mathscr{B}_{I}\otimes\mathscr{F}$, and is the $\sigma$-algebra generated by the collection of functions $I\times\Omega\to E$ that are progressively measurable with respect to the filtration $\mathscr{F}_{I}$:

 $\mathrm{Prog}(\mathscr{F}_{I})=\sigma(\mathscr{M}^{0}(\mathscr{F}_{I})).$
2. 2.

If $X:I\times\Omega\to E$ is progressively measurable with respect to the filtration $\mathscr{F}_{I}$, then the stochastic process $(X_{t})_{t\in I}$ is jointly measurable and is adapted to the filtration.

###### Proof.

If $A_{1},A_{2},\ldots\in\mathrm{Prog}(\mathscr{F}_{I})$ and $t\in I$ then

 $([0,t]\times\Omega)\cap\bigcup_{n\geq 1}A_{n}=\bigcup_{n\geq 1}(([0,t]\times% \Omega)\cap A_{n}),$

which is a countable union of elements of the $\sigma$-algebra $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$ and hence belongs to $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$, showing that $\bigcup_{n\geq 1}A_{n}\in\mathrm{Prog}(\mathscr{F}_{I})$. If $A_{1},A_{2}\in\mathrm{Prog}(\mathscr{F}_{I})$ and $t\in I$ then

 $([0,t]\times\Omega)\cap(A_{1}\cap A_{2})=(([0,t]\times\Omega)\cap A_{1})\cap((% [0,t]\times\Omega)\cap A_{2}),$

which is an intersection of two elements of $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$ and hence belongs to $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$, showing that $A_{1}\cap A_{2}\in\mathrm{Prog}(\mathscr{F}_{I})$. Thus $\mathrm{Prog}((\mathscr{F}_{t})_{t\in I})$ is a $\sigma$-algebra.

If $X:I\times\Omega\to E$ is progressively measurable, $B\in\mathscr{E}$, and $t\in I$, then

 $([0,t]\times\Omega)\cap X^{-1}(B)=\{(s,\omega)\in[0,t]\times\Omega:X(s,\omega)% \in B\}.$

Because $X$ is progressively measurable, this belongs to $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$. This is true for all $t$, hence $X^{-1}(B)\in\mathrm{Prog}(\mathscr{F}_{I})$, which means that $X$ is measurable $\mathrm{Prog}(\mathscr{F}_{I})\to\mathscr{E}$.

If $X:I\times\Omega\to E$ is measurable $\mathrm{Prog}(\mathscr{F}_{I})\to\mathscr{E}$, $t\in I$, and $B\in\mathscr{E}$, then because $X^{-1}(B)\in\mathrm{Prog}(\mathscr{F}_{I})$, we have $([0,t]\times\Omega)\cap X^{-1}(B)\in\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$. This is true for all $B\in\mathscr{E}$, which means that $(s,\omega)\mapsto X(s,\omega)$, $[0,t]\times\Omega\to E$, is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$, and because this is true for all $t$, $X$ is progressively measurable. Therefore a function $I\times\Omega\to E$ is progressively measurable if and only if it is measurable $\mathrm{Prog}(\mathscr{F}_{I})\to\mathscr{E}$, which shows that $\mathrm{Prog}(\mathscr{F}_{I})$ is the coarsest $\sigma$-algebra with which all progressively measurable functions are measurable.

If $X:I\times\Omega\to E$ is a progressively measurable function and $B\in\mathscr{E}$,

 $X^{-1}(B)=\bigcup_{k\geq 1}(([0,k]\times\Omega)\cap X^{-1}(B)).$

Because $X$ is progressively measurable,

 $([0,k]\times\Omega)\cap X^{-1}(B)\in\mathscr{B}_{[0,k]}\otimes\mathscr{F}_{k}% \subset\mathscr{B}_{I}\otimes\mathscr{F},$

thus $X^{-1}(B)$ is equal to a countable union of elements of $\mathscr{B}_{I}\otimes\mathscr{F}$ and so itself belongs to $\mathscr{B}_{I}\otimes\mathscr{F}$. Therefore $X$ is measurable $\mathscr{B}_{I}\otimes\mathscr{F}\to\mathscr{E}$, namely $X$ is jointly measurable.

Because $\mathrm{Prog}(\mathscr{F}_{I})$ is the $\sigma$-algebra generated by the collection of progressively measurable functions and each progressively measurable function is measurable $\mathscr{B}_{I}\otimes\mathscr{F}$,

 $\mathrm{Prog}(\mathscr{F}_{I})\subset\mathscr{B}_{I}\otimes\mathscr{F},$

and so $\mathrm{Prog}(\mathscr{F}_{I})$ is indeed a $\sigma$-subalgebra of $\mathscr{B}_{I}\otimes\mathscr{F}$.

Let $t\in I$. That $X$ is progressively measurable means that

 $(s,\omega)\mapsto X(s,\omega),\qquad[0,t]\times\Omega$

is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$. This implies that for each $s\in[0,t]$ the map $\omega\mapsto X(s,\omega)$ is measurable $\mathscr{F}_{t}\to\mathscr{E}$.33 3 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 152, Theorem 4.48. (Generally, if a function is jointly measurable then it is separately measurable in each argument.) In particular, $\omega\mapsto X(t,\omega)$ is measurable $\mathscr{F}_{t}\to\mathscr{E}$, which means that the stochastic process $(X_{t})_{t\in I}$ is adapted to the filtration, completing the proof. ∎

We now prove that if a stochastic process is adapted and left-continuous then it is progressively measurable.44 4 cf. Daniel W. Stroock, Probability Theory: An Analytic View, second ed., p. 267, Lemma 7.1.2.

###### Theorem 3.

Let $(\mathscr{F}_{t})_{t\in I}$ be a filtration of $\mathscr{F}$. If $(X_{t})_{t\in I}$ is a stochastic process that is adapted to this filtration and all its paths are left-continuous, then $X$ is progressively measurable with respect to this filtration.

###### Proof.

Write $X(t,\omega)=X_{t}(\omega)$. For $t\in I$, let $Y$ be the restriction of $X$ to $[0,t]\times\Omega$. We wish to prove that $Y$ is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$. For $n\geq 1$, define

 $Y_{n}(s,\omega)=\sum_{k=0}^{2^{n}-1}1_{[kt2^{-n},(k+1)t2^{-n})}(s)Y(kt2^{-n},% \omega)+1_{\{t\}}(s)Y(t,\omega).$

Because $X$ is adapted to the filtration, each $Y_{n}$ is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$. Because $X$ has left-continuous paths, for $(s,\omega)\in[0,t]\times\Omega$,

 $Y_{n}(s,\omega)\to Y(s,\omega).$

Since $Y$ is the pointwise limit of $Y_{n}$, it follows that $Y$ is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$, and so $X$ is progressively measurable. ∎

## 4 Stopping times

Let $\mathscr{F}_{I}=(\mathscr{F}_{t})_{t\in I}$ be a filtration of $\mathscr{F}$. A function $T:\Omega\to[0,\infty]$ is called a stopping time with respect to the filtration $\mathscr{F}_{I}$ if

 $\{T\leq t\}\in\mathscr{F}_{t},\qquad t\in I.$

It is straightforward to prove that a stopping time is measurable $\mathscr{F}\to\mathscr{B}_{[0,\infty]}$. Let

 $\mathscr{F}_{\infty}=\sigma(\mathscr{F}_{t}:t\in I).$

We define

 $\mathscr{F}_{T}=\{A\in\mathscr{F}_{\infty}:\textrm{if t\in I then A\cap\{T% \leq t\}\in\mathscr{F}_{t}}\}.$

It is straightforward to check that $T$ is measurable $\mathscr{F}_{T}\to\mathscr{B}_{[0,\infty]}$, and in particular $\{T<\infty\}\in\mathscr{F}_{T}$.

For a stochastic process $(X_{t})_{t\in I}$ with state space $E$, we define $X_{T}:\Omega\to E$ by

 $X_{T}(\omega)=1_{\{T<\infty\}}(\omega)X_{T(\omega)}(\omega).$

We prove that if $X$ is progressively measurable then $X_{T}$ is measurable $\mathscr{F}_{T}\to\mathscr{E}$.55 5 Sheng-wu He and Jia-gang Wang and Jia-An Yan, Semimartingale Theory and Stochastic Calculus, p. 86, Theorem 3.12.

###### Theorem 4.

If $\mathscr{F}_{I}=(\mathscr{F}_{t})_{t\in I}$ is a filtration of $\mathscr{F}$, $(X_{t})_{t\in I}$ is a stochastic process that is progressively measurable with respect to $\mathscr{F}_{I}$, and $T$ is a stopping time with respect to $\mathscr{F}_{I}$, then $X_{T}$ is measurable $\mathscr{F}_{T}\to\mathscr{E}$.

###### Proof.

For $t\in I$, using that $T$ is a stopping time we check that $\omega\mapsto T(\omega)\wedge t$ is measurable $\mathscr{F}_{t}\to\mathscr{B}_{[0,t]}$, and then $\omega\mapsto(T(\omega)\wedge t,\omega)$, $\Omega\to[0,t]\times\Omega$, is measurable $\mathscr{F}_{t}\to\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}$.66 6 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 152, Lemma 4.49. Because $X$ is progressively measurable, $(s,\omega)\mapsto X_{s}(\omega)$ is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$. Therefore the composition

 $\omega\mapsto X_{T(\omega)\wedge t}(\omega),\qquad\Omega\to E,$

is measurable $\mathscr{F}_{t}\to\mathscr{E}$, and a fortiori it is measurable $\mathscr{F}_{\infty}\to\mathscr{E}$. We have

 $X_{T}(\omega)=\lim_{n\to\infty}1_{\{T\leq n\}}(\omega)X_{T(\omega)\wedge n}(% \omega),$

and because $\omega\mapsto 1_{\{T\leq n\}}(\omega)X_{T(\omega)\wedge n}(\omega)$ is measurable $\mathscr{F}_{\infty}\to\mathscr{E}$, it follows that $\omega\mapsto X_{T}(\omega)$ is measurable $\mathscr{F}_{\infty}\to\mathscr{E}$. For $B\in\mathscr{E}$,

 $\{X_{T}\in B\}\cap\{T\leq t\}=\{\omega\in\Omega:X_{T(\omega)\wedge t}(\omega)% \in B\}\cap\{T\leq t\}\in\mathscr{F}_{t},$

therefore $\{X_{T}\in B\}\in\mathscr{F}_{T}$. This means that $X_{T}$ is measurable $\mathscr{F}_{T}\to\mathscr{E}$. ∎

For a stochastic process $(X_{t})_{t\in I}$, a filtration $\mathscr{F}_{I}=(\mathscr{F}_{t})_{t\in I}$, and a stopping time $T$ with respect to the filtration, we define

 $X^{T}_{t}(\omega)=X_{T(\omega)\wedge t}(\omega),$

and $(X^{T}_{t})_{t\in I}$ is a stochastic process. We prove that if $X$ is progressively measurable with respect to $\mathscr{F}_{I}$ then the stochastic proces $X^{T}$ is progressively measurable with respect to $\mathscr{F}_{I}$.77 7 Ioannis Karatzas and Steven Shreve, Brownian Motion and Stochastic Calculus, p. 9, Proposition 2.18.

###### Theorem 5.

If $(X_{t})_{t\in I}$ is a stochastic process that is progressively measurable with respect to a filtration $\mathscr{F}_{I}=(\mathscr{F}_{t})_{t\in I}$ and $T$ is a stopping time with respect to $\mathscr{F}_{I}$, then $X^{T}$ is progressively measurable with respect to $\mathscr{F}_{I}$.

###### Proof.

Let $t\in I$. Because $T$ is a stopping time, for each $s\in[0,t]$ the map $\omega\mapsto T(\omega)\wedge s$ is measurable $\mathscr{F}_{s}\to\mathscr{B}_{[0,t]}$ and a fortiori is measurable $\mathscr{F}_{t}\to\mathscr{B}_{[0,t]}$. Therefore $(s,\omega)\mapsto T(\omega)\wedge s$ is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{B}_{[0,t]}$,88 8 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 152, Theorem 4.48. and $(s,\omega)\mapsto\omega$ is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{F}_{t}$. This implies that

 $(s,\omega)\mapsto(T(\omega)\wedge s,\omega),\qquad[0,t]\times\Omega\to[0,t]% \times\Omega,$

is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{B}_{[0,t]}\otimes\mathscr% {F}_{t}$.99 9 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 152, Lemma 4.49. Because $X$ is progressively measurable,

 $(s,\omega)\mapsto X_{s}(\omega),\qquad[0,t]\times\Omega\to E,$

is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$. Therefore the composition

 $(s,\omega)\mapsto X_{T(\omega)\wedge s}(\omega),\qquad[0,t]\times\Omega\to E,$

is measurable $\mathscr{B}_{[0,t]}\otimes\mathscr{F}_{t}\to\mathscr{E}$, which shows that $X^{T}$ is progressively measurable. ∎