Is this proof of measurability correct?

38 Views Asked by At

Let $(X, \mathcal X)$, $(Y, \mathcal Y)$, and $(Z, \mathcal Z)$ be measurable spaces. Let $\mathcal X'$ be a sub-algebra of $\mathcal X$.

Let $a: X \times Y \to Z$ be $(\mathcal X \otimes \mathcal Y, \mathcal Z)$ measurable.

Let $b: X \to Y$ be $(\mathcal X', \mathcal Y)$ measurable.

Is the function $c: X \times X \to Z$ defined by $$c(x,x') = a(x,b(x'))$$ $(\mathcal X \otimes \mathcal X', \mathcal Z)$ measurable?

Attempt. First, let $d: (x,x') \mapsto (x, b(x'))$. Since $\pi_1 \circ d = \pi_1$ and $\pi_2 \circ d = b \circ \pi_2$, $d$ is $(\mathcal X \otimes \mathcal X', \mathcal X \otimes \mathcal Y)$ measurable (where $\pi_i$ is the projection onto the $i$th coordinate). Then, $c = a \circ d$. So $c$ is $(\mathcal X \otimes \mathcal X', \mathcal Z)$ measurable.