Time dependent process and progressively measurability

Question

I'm trying to show that

$f : \left[0,\infty\right)\times\mathbb{R}\rightarrow\mathbb{R}\text{ is a Borel function},X : \Omega\times\left[0,\infty\right)\text{ is a progressively measurable process}\Rightarrow \left( f \left( t,X_t \right) \right)_{t\geq0} : \text{is progressively measurable under naturally given filtered space.}$

I think if $g\left(\omega,t \right)=\left(t,X_t\right)$ is progressively measurable, then the other part is clear. But I have no idea to deal with inverse images of $X$ and $g$.

Am I doing right? If not, how do I have to approach?

Answer 1

Claim 1: $(t,X_t)$ is progressively measurable. This follows from the following lemma:

Let $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ be two real-valued stochastic processes on a probabiltiy space $(\Omega,\mathcal{A})$ with filtration $(\mathcal{F}_t)_{t \geq 0}$. If $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are progressively measurable, then also $$(t,\omega) \mapsto Z_t(\omega) := (X_t(\omega),Y_t(\omega)) \in \mathbb{R}^2$$ is progressively measurable.

Proof: Fix $T \geq 0$. If $B_1,B_2 \in \mathcal{B}(\mathbb{R})$ are Borel sets, then it follows from the progressive measurability of $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ that $$\{(t,\omega) \in [0,T] \times \Omega; X_t(\omega) \in B_1\} \in \mathcal{F}_T \quad \text{and} \quad \{(t,\omega) \in [0,T] \times \Omega; Y_t(\omega) \in B_2\} \in \mathcal{F}_T.$$ Thus, \begin{align*} &\{(t,\omega) \in [0,T] \times \Omega; Z_t(\omega) \in B_1\times B_2\} \\ &= \{(t,\omega) \in [0,T] \times \Omega; X_t(\omega) \in B_1\} \cap \{(t,\omega) \in [0,T] \times \Omega; Y_t(\omega) \in B_2\} \\ &\in \mathcal{F}_T.\end{align*} Since sets of the form $B_1 \times B_2$ with $B_i \in \mathcal{B}(\mathbb{R})$ are a generator of $\mathcal{B}(\mathbb{R}^2)$, this implies $$\{(t,\omega) \in [0,T] \times \Omega; Z_t(\omega) \in B\} \in \mathcal{F}_T$$ for any $B \in \mathcal{B}(\mathbb{R}^2)$, i.e. $Z$ is progressively measurable.

Claim 2: If $(Z_t)_{t \geq 0}$ is a realvalued progressively measurable process and $f: \mathbb{R} \to \mathbb{R}$ a Borel measurable function, then $(f(Z_t))_{t \geq 0}$ is progressively measurable.

Proof: Fix $T \geq 0$. For any Borel set $B \in \mathcal{B}(\mathbb{R})$ we have $$\{(t,\omega) \in [0,T] \times \Omega; f(Z_t(\omega)) \in B\} = \{(t,\omega) \in [0,T] \times \Omega; Z_t(\omega)) \in f^{-1}(B)\}.$$ Since $(Z_t)_{t \geq 0}$ is progressively measurable and $f^{-1}(B) \in \mathcal{B}(\mathbb{R})$, it follows immediately that the right-hand side (and hence the left-hand side) is in $\mathcal{F}_T$. This means that $(f(Z_t))_{t \geq 0}$ is progressively measurable.