Jointly measurable and progressively measurable stochastic processes
1 Jointly measurable stochastic processes
Let $E={\mathbb{R}}^{d}$ with Borel $\mathcal{E}$, let $I={\mathbb{R}}_{\ge 0}$, which is a topological space with the subspace topology inherited from $\mathbb{R}$, and let $(\mathrm{\Omega},\mathcal{F},P)$ 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 ${\mathcal{B}}_{I}\otimes \mathcal{F}\to \mathcal{E}$.
For $\omega \in \mathrm{\Omega}$, the path $t\mapsto {X}_{t}(\omega )$ is called leftcontinuous if for each $t\in I$,
$${X}_{s}(\omega )\to {X}_{t}(\omega ),s\uparrow t.$$ 
We prove that if the paths of a stochastic process are leftcontinuous then the stochastic process is jointly measurable.^{1}^{1} 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 leftcontinuous, then $X$ is jointly measurable.
Proof.
For $n\ge 1$, $t\in I$, and $\omega \in \mathrm{\Omega}$, let
$${X}_{t}^{n}(\omega )=\sum _{k=0}^{\mathrm{\infty}}{1}_{[k{2}^{n},(k+1){2}^{n})}(t){X}_{k{2}^{n}}(\omega ).$$ 
Each ${X}^{n}$ is measurable ${\mathcal{B}}_{I}\otimes \mathcal{F}\to \mathcal{E}$: for $B\in \mathcal{E}$,
$$\{(t,\omega )\in I\times \mathrm{\Omega}:{X}_{t}^{n}(\omega )\in B\}=\bigcup _{k=0}^{\mathrm{\infty}}[k{2}^{n},(k+1){2}^{n})\times \{{X}_{k{2}^{n}}\in B\}.$$ 
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 leftcontinuous, ${X}_{{k}_{n}{2}^{n}}(\omega )\to {X}_{t}(\omega )$. That is, ${X}^{n}\to X$ pointwise on $I\times \mathrm{\Omega}$, and because each ${X}_{n}$ is measurable ${\mathcal{B}}_{I}\otimes \mathcal{F}\to \mathcal{E}$ this implies that $X$ is measurable ${\mathcal{B}}_{I}\otimes \mathcal{F}\to \mathcal{E}$.^{2}^{2} 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. ∎
2 Adapted stochastic processes
Let ${\mathcal{F}}_{I}={({\mathcal{F}}_{t})}_{t\in I}$ be a filtration of $\mathcal{F}$. A stochastic process $X$ is said to be adapted to the filtration ${\mathrm{F}}_{I}$ if for each $t\in I$ the map
$$\omega \mapsto {X}_{t}(\omega ),\mathrm{\Omega}\to E,$$ 
is measurable ${\mathcal{F}}_{t}\to \mathcal{E}$, in other words, for each $t\in I$,
$$\sigma ({X}_{t})\subset {\mathcal{F}}_{t}.$$ 
For a stochastic process ${({X}_{t})}_{t\in I}$, the natural filtration of $X$ is
$$\sigma ({X}_{s}:s\le t).$$ 
It is immediate that this is a filtration and that $X$ is adapted to it.
3 Progressively measurable stochastic processes
Let ${\mathcal{F}}_{I}={({\mathcal{F}}_{t})}_{t\in I}$ be a filtration of $\mathcal{F}$. A function $X:I\times \mathrm{\Omega}\to E$ is called progressively measurable with respect to the filtration ${\mathrm{F}}_{I}$ if for each $t\in I$, the map
$$(s,\omega )\mapsto X(s,\omega ),[0,t]\times \mathrm{\Omega}\to E,$$ 
is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$. We denote by ${\mathcal{M}}^{0}({\mathcal{F}}_{I})$ the set of functions $I\times \mathrm{\Omega}\to E$ that are progressively measurable with respect to the filtration ${\mathcal{F}}_{I}$. 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 $\mathrm{Prog}({\mathcal{F}}_{I})$ the collection of those subsets $A$ of $I\times \mathrm{\Omega}$ such that for each $t\in I$,
$$([0,t]\times \mathrm{\Omega})\cap A\in {\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}.$$ 
We prove in the following that this is a $\sigma $subalgebra of ${\mathcal{B}}_{I}\otimes \mathcal{F}$ and that it is the coarsest $\sigma $algebra with which all progressively measurable functions are measurable.
Theorem 2.
Let ${\mathcal{F}}_{I}={({\mathcal{F}}_{t})}_{t\in I}$ be a filtration of $\mathcal{F}$.

1.
$\mathrm{Prog}({\mathcal{F}}_{I})$ is a $\sigma $subalgebra of ${\mathcal{B}}_{I}\otimes \mathcal{F}$, and is the $\sigma $algebra generated by the collection of functions $I\times \mathrm{\Omega}\to E$ that are progressively measurable with respect to the filtration ${\mathcal{F}}_{I}$:
$$\mathrm{Prog}({\mathcal{F}}_{I})=\sigma ({\mathcal{M}}^{0}({\mathcal{F}}_{I})).$$ 
2.
If $X:I\times \mathrm{\Omega}\to E$ is progressively measurable with respect to the filtration ${\mathcal{F}}_{I}$, then the stochastic process ${({X}_{t})}_{t\in I}$ is jointly measurable and is adapted to the filtration.
Proof.
If ${A}_{1},{A}_{2},\mathrm{\dots}\in \mathrm{Prog}({\mathcal{F}}_{I})$ and $t\in I$ then
$$([0,t]\times \mathrm{\Omega})\cap \bigcup _{n\ge 1}{A}_{n}=\bigcup _{n\ge 1}(([0,t]\times \mathrm{\Omega})\cap {A}_{n}),$$ 
which is a countable union of elements of the $\sigma $algebra ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$ and hence belongs to ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$, showing that ${\bigcup}_{n\ge 1}{A}_{n}\in \mathrm{Prog}({\mathcal{F}}_{I})$. If ${A}_{1},{A}_{2}\in \mathrm{Prog}({\mathcal{F}}_{I})$ and $t\in I$ then
$$([0,t]\times \mathrm{\Omega})\cap ({A}_{1}\cap {A}_{2})=(([0,t]\times \mathrm{\Omega})\cap {A}_{1})\cap (([0,t]\times \mathrm{\Omega})\cap {A}_{2}),$$ 
which is an intersection of two elements of ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$ and hence belongs to ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$, showing that ${A}_{1}\cap {A}_{2}\in \mathrm{Prog}({\mathcal{F}}_{I})$. Thus $\mathrm{Prog}({({\mathcal{F}}_{t})}_{t\in I})$ is a $\sigma $algebra.
If $X:I\times \mathrm{\Omega}\to E$ is progressively measurable, $B\in \mathcal{E}$, and $t\in I$, then
$$([0,t]\times \mathrm{\Omega})\cap {X}^{1}(B)=\{(s,\omega )\in [0,t]\times \mathrm{\Omega}:X(s,\omega )\in B\}.$$ 
Because $X$ is progressively measurable, this belongs to ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$. This is true for all $t$, hence ${X}^{1}(B)\in \mathrm{Prog}({\mathcal{F}}_{I})$, which means that $X$ is measurable $\mathrm{Prog}({\mathcal{F}}_{I})\to \mathcal{E}$.
If $X:I\times \mathrm{\Omega}\to E$ is measurable $\mathrm{Prog}({\mathcal{F}}_{I})\to \mathcal{E}$, $t\in I$, and $B\in \mathcal{E}$, then because ${X}^{1}(B)\in \mathrm{Prog}({\mathcal{F}}_{I})$, we have $([0,t]\times \mathrm{\Omega})\cap {X}^{1}(B)\in {\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$. This is true for all $B\in \mathcal{E}$, which means that $(s,\omega )\mapsto X(s,\omega )$, $[0,t]\times \mathrm{\Omega}\to E$, is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$, and because this is true for all $t$, $X$ is progressively measurable. Therefore a function $I\times \mathrm{\Omega}\to E$ is progressively measurable if and only if it is measurable $\mathrm{Prog}({\mathcal{F}}_{I})\to \mathcal{E}$, which shows that $\mathrm{Prog}({\mathcal{F}}_{I})$ is the coarsest $\sigma $algebra with which all progressively measurable functions are measurable.
If $X:I\times \mathrm{\Omega}\to E$ is a progressively measurable function and $B\in \mathcal{E}$,
$${X}^{1}(B)=\bigcup _{k\ge 1}(([0,k]\times \mathrm{\Omega})\cap {X}^{1}(B)).$$ 
Because $X$ is progressively measurable,
$$([0,k]\times \mathrm{\Omega})\cap {X}^{1}(B)\in {\mathcal{B}}_{[0,k]}\otimes {\mathcal{F}}_{k}\subset {\mathcal{B}}_{I}\otimes \mathcal{F},$$ 
thus ${X}^{1}(B)$ is equal to a countable union of elements of ${\mathcal{B}}_{I}\otimes \mathcal{F}$ and so itself belongs to ${\mathcal{B}}_{I}\otimes \mathcal{F}$. Therefore $X$ is measurable ${\mathcal{B}}_{I}\otimes \mathcal{F}\to \mathcal{E}$, namely $X$ is jointly measurable.
Because $\mathrm{Prog}({\mathcal{F}}_{I})$ is the $\sigma $algebra generated by the collection of progressively measurable functions and each progressively measurable function is measurable ${\mathcal{B}}_{I}\otimes \mathcal{F}$,
$$\mathrm{Prog}({\mathcal{F}}_{I})\subset {\mathcal{B}}_{I}\otimes \mathcal{F},$$ 
and so $\mathrm{Prog}({\mathcal{F}}_{I})$ is indeed a $\sigma $subalgebra of ${\mathcal{B}}_{I}\otimes \mathcal{F}$.
Let $t\in I$. That $X$ is progressively measurable means that
$$(s,\omega )\mapsto X(s,\omega ),[0,t]\times \mathrm{\Omega}$$ 
is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$. This implies that for each $s\in [0,t]$ the map $\omega \mapsto X(s,\omega )$ is measurable ${\mathcal{F}}_{t}\to \mathcal{E}$.^{3}^{3} 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 ${\mathcal{F}}_{t}\to \mathcal{E}$, 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 leftcontinuous then it is progressively measurable.^{4}^{4} 4 cf. Daniel W. Stroock, Probability Theory: An Analytic View, second ed., p. 267, Lemma 7.1.2.
Theorem 3.
Let ${({\mathcal{F}}_{t})}_{t\in I}$ be a filtration of $\mathcal{F}$. If ${({X}_{t})}_{t\in I}$ is a stochastic process that is adapted to this filtration and all its paths are leftcontinuous, 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 $[0,t]\times \mathrm{\Omega}$. We wish to prove that $Y$ is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$. For $n\ge 1$, define
$${Y}_{n}(s,\omega )=\sum _{k=0}^{{2}^{n}1}{1}_{[kt{2}^{n},(k+1)t{2}^{n})}(s)Y(kt{2}^{n},\omega )+{1}_{\{t\}}(s)Y(t,\omega ).$$ 
Because $X$ is adapted to the filtration, each ${Y}_{n}$ is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$. Because $X$ has leftcontinuous paths, for $(s,\omega )\in [0,t]\times \mathrm{\Omega}$,
$${Y}_{n}(s,\omega )\to Y(s,\omega ).$$ 
Since $Y$ is the pointwise limit of ${Y}_{n}$, it follows that $Y$ is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$, and so $X$ is progressively measurable. ∎
4 Stopping times
Let ${\mathcal{F}}_{I}={({\mathcal{F}}_{t})}_{t\in I}$ be a filtration of $\mathcal{F}$. A function $T:\mathrm{\Omega}\to [0,\mathrm{\infty}]$ is called a stopping time with respect to the filtration ${\mathrm{F}}_{I}$ if
$$\{T\le t\}\in {\mathcal{F}}_{t},t\in I.$$ 
It is straightforward to prove that a stopping time is measurable $\mathcal{F}\to {\mathcal{B}}_{[0,\mathrm{\infty}]}$. Let
$${\mathcal{F}}_{\mathrm{\infty}}=\sigma ({\mathcal{F}}_{t}:t\in I).$$ 
We define
$${\mathcal{F}}_{T}=\{A\in {\mathcal{F}}_{\mathrm{\infty}}:\text{if}t\in I\text{then}A\cap \{T\le t\}\in {\mathcal{F}}_{t}\}.$$ 
It is straightforward to check that $T$ is measurable ${\mathcal{F}}_{T}\to {\mathcal{B}}_{[0,\mathrm{\infty}]}$, and in particular $$.
For a stochastic process ${({X}_{t})}_{t\in I}$ with state space $E$, we define ${X}_{T}:\mathrm{\Omega}\to E$ by
$$ 
We prove that if $X$ is progressively measurable then ${X}_{T}$ is measurable ${\mathcal{F}}_{T}\to \mathcal{E}$.^{5}^{5} 5 Shengwu He and Jiagang Wang and JiaAn Yan, Semimartingale Theory and Stochastic Calculus, p. 86, Theorem 3.12.
Theorem 4.
If ${\mathcal{F}}_{I}={({\mathcal{F}}_{t})}_{t\in I}$ is a filtration of $\mathcal{F}$, ${({X}_{t})}_{t\in I}$ is a stochastic process that is progressively measurable with respect to ${\mathcal{F}}_{I}$, and $T$ is a stopping time with respect to ${\mathcal{F}}_{I}$, then ${X}_{T}$ is measurable ${\mathcal{F}}_{T}\to \mathcal{E}$.
Proof.
For $t\in I$, using that $T$ is a stopping time we check that $\omega \mapsto T(\omega )\wedge t$ is measurable ${\mathcal{F}}_{t}\to {\mathcal{B}}_{[0,t]}$, and then $\omega \mapsto (T(\omega )\wedge t,\omega )$, $\mathrm{\Omega}\to [0,t]\times \mathrm{\Omega}$, is measurable ${\mathcal{F}}_{t}\to {\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$.^{6}^{6} 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 ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$. Therefore the composition
$$\omega \mapsto {X}_{T(\omega )\wedge t}(\omega ),\mathrm{\Omega}\to E,$$ 
is measurable ${\mathcal{F}}_{t}\to \mathcal{E}$, and a fortiori it is measurable ${\mathcal{F}}_{\mathrm{\infty}}\to \mathcal{E}$. We have
$${X}_{T}(\omega )=\underset{n\to \mathrm{\infty}}{lim}{1}_{\{T\le n\}}(\omega ){X}_{T(\omega )\wedge n}(\omega ),$$ 
and because $\omega \mapsto {1}_{\{T\le n\}}(\omega ){X}_{T(\omega )\wedge n}(\omega )$ is measurable ${\mathcal{F}}_{\mathrm{\infty}}\to \mathcal{E}$, it follows that $\omega \mapsto {X}_{T}(\omega )$ is measurable ${\mathcal{F}}_{\mathrm{\infty}}\to \mathcal{E}$. For $B\in \mathcal{E}$,
$$\{{X}_{T}\in B\}\cap \{T\le t\}=\{\omega \in \mathrm{\Omega}:{X}_{T(\omega )\wedge t}(\omega )\in B\}\cap \{T\le t\}\in {\mathcal{F}}_{t},$$ 
therefore $\{{X}_{T}\in B\}\in {\mathcal{F}}_{T}$. This means that ${X}_{T}$ is measurable ${\mathcal{F}}_{T}\to \mathcal{E}$. ∎
For a stochastic process ${({X}_{t})}_{t\in I}$, a filtration ${\mathcal{F}}_{I}={({\mathcal{F}}_{t})}_{t\in I}$, 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 ${\mathcal{F}}_{I}$ then the stochastic proces ${X}^{T}$ is progressively measurable with respect to ${\mathcal{F}}_{I}$.^{7}^{7} 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 ${\mathcal{F}}_{I}={({\mathcal{F}}_{t})}_{t\in I}$ and $T$ is a stopping time with respect to ${\mathcal{F}}_{I}$, then ${X}^{T}$ is progressively measurable with respect to ${\mathcal{F}}_{I}$.
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 ${\mathcal{F}}_{s}\to {\mathcal{B}}_{[0,t]}$ and a fortiori is measurable ${\mathcal{F}}_{t}\to {\mathcal{B}}_{[0,t]}$. Therefore $(s,\omega )\mapsto T(\omega )\wedge s$ is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to {\mathcal{B}}_{[0,t]}$,^{8}^{8} 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 ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to {\mathcal{F}}_{t}$. This implies that
$$(s,\omega )\mapsto (T(\omega )\wedge s,\omega ),[0,t]\times \mathrm{\Omega}\to [0,t]\times \mathrm{\Omega},$$ 
is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to {\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}$.^{9}^{9} 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,
$$(s,\omega )\mapsto {X}_{s}(\omega ),[0,t]\times \mathrm{\Omega}\to E,$$ 
is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$. Therefore the composition
$$(s,\omega )\mapsto {X}_{T(\omega )\wedge s}(\omega ),[0,t]\times \mathrm{\Omega}\to E,$$ 
is measurable ${\mathcal{B}}_{[0,t]}\otimes {\mathcal{F}}_{t}\to \mathcal{E}$, which shows that ${X}^{T}$ is progressively measurable. ∎