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Consider the following problem:

  1. Let $\mathbb{K} = \mathbb{R}$ or $\mathbb{K} = \mathbb{C}$.
  2. Assume that $a \in \mathbb{R}$ and the function $F:[a, \infty) \times \mathbb{K} \rightarrow \mathbb{K}$ is Borel measurable.
  3. Let $X:[a, \infty) \rightarrow \mathbb{K}$ be Borel measurable.

Show that the mapping $[a, \infty) \rightarrow \mathbb{K}$, $s \mapsto F(s, X(s))$ is Borel measurable.


First of all, I understand the Borel measurability of $F$ in the following way: $\forall B \in \mathscr{B}(K)$ $$ {F(s,z)}^{-1}(B) = \{ (s,z)\in [a, \infty) \times \mathbb{K}: F(s,z)\in B\} \in \mathscr{B}([a,\infty)) \otimes \mathscr{B}(K). $$

I can show that for every fixed $z'$ the map $[a, \infty) \rightarrow \mathbb{K}$, $s \mapsto F(s, z')$ is Borel measurable using the following result.

Definition:

Let $(\Omega_1, \mathcal{F}_1)$ and $(\Omega_2, \mathcal{F}_2)$ be two measurable spaces. For every $Q \subset \Omega_1 \times \Omega_2$, and $\omega_1 \in \Omega_1, \omega_2 \in \Omega_2$ the sets \begin{align*} Q_{\omega_1} & := \{ w_2 \in \Omega_2: (\omega_1, \omega_2) \in Q \}, \\ Q_{\omega_2} & := \{ w_1 \in \Omega_1: (\omega_1, \omega_2) \in Q \} \end{align*} are called, respectively, the $\omega_1$-section and the $\omega_2$-section of $Q$.

Lemma:

Let $(\Omega_1, \mathcal{F}_1)$ and $(\Omega_2, \mathcal{F}_2)$ be two measurable spaces. If $Q \in \mathcal{F}_1 \otimes \mathcal{F}_2$, then

  • for every $\omega_1 \in \Omega_1$ $Q_{\omega_1} \in \mathcal{F}_2$,
  • for every $\omega_2 \in \Omega_2$ $Q_{\omega_2} \in \mathcal{F}_1$.

Thus, to show that $F(s,z')$ is Borel measurable, write

$$ {F(s,z')}^{-1}(B) = \{ s \in [a, \infty): F(s,z')\in B\} = { \{ (s,z) \in [a, \infty) \times \mathbb{K}: F(s,z)\in B\} }_{z'} \in \mathscr{B}([a,\infty)). $$

In a similar way, one can show that $F(s',z)$ is Borel measurable for a fixed $s'$.

Now, how can I show this for $F(s,X(s))$?

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1 Answer 1

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The function $G:[a,\infty)\to[a,\infty)\times\mathbb K$ prescribed by $s\mapsto\langle s,X(s)\rangle$ is Borel measurable.

This because the compositions $\mathsf{id}=\pi_1\circ G:[a,\infty)\to[a,\infty)$ and $X=\pi_2\circ G:[a,\infty)\to\mathbb K$ - where the $\pi_i$ denote the projections - are both Borel measurable.

That means at first hand that $G$ is measurable if $[a,\infty)\times\mathbb K$ is equipped with $\sigma$-algebra $\mathscr B([a,\infty))\otimes\mathscr B(\mathbb K)$.

But it can be proved that: $$\mathscr B([a,\infty))\otimes\mathscr B(\mathbb K)=\mathscr B([a,\infty)\times\mathbb K)$$

So $G$ is also Borel measurable.

Then $F\circ G$ is a composition of Borel measurable functions, hence is Borel measurable.

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  • $\begingroup$ Am I right in assuming that the following fact is used? Let $(\Omega_1, \mathcal{F}_1)$, $(\Omega_2, \mathcal{F}_2)$ and $(\Omega, \mathcal{F})$ be measurable spaces. Then $G : \Omega \rightarrow \Omega_1 \times \Omega_2$ is $\mathcal{F} - \mathcal{F}_1 \otimes \mathcal{F}_2 $ measurable if and only if $\pi_j \circ G$ is $\mathcal{F}-\mathcal{F}_j$ measurable $(j = 1,2)$. $\endgroup$
    – Holden
    Nov 29, 2017 at 10:50
  • $\begingroup$ Yes, that is applied here. $\endgroup$
    – drhab
    Nov 29, 2017 at 10:51

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