Exercise 0.18 (Joint functions into product spaces preserve measurability)

We verify Definition 16. Let $F$ be the $\sigma$-algebra of $\mathrm{\Omega }$. By Example 14, let $B={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{B}_i$ be the product $\sigma$-algebra of $R_1\times \dots \times R_n$.

We make use of the measure-theoretic induction ( Exercise 11) again. The generating set of $B$ is:

Let $S$ be an arbitrary set in the above collection. By definition, for some $1\le j\le n$ the elements of $S$ contain all of the combinations of the elements of some fixed $E\in B_j$ on the $j$-th position with the whole spaces $R_1,\dots ,R_{j-1},R_{j+1},\dots ,R_n$ on the other positions. Therefore, we can use how Cartesian products interact with inverse images (add source) to derive that

Since $X_j$ is a measurable morphism and $E\in B_j$, $X_j^{-1}(E)$ must be measurable in $(\mathrm{\Omega },F)$.

Now we demonstrate that $P(S)=X^{-1}(S)\in F$ is a $\sigma$-algebra property.

• We have $P(\varnothing )$ is true, since the inverse image of the empty set is always the empty set itself, whereas the empty set is an element of any $\sigma$-algebra.
• Suppose that $X^{-1}(S)\in F$. By Exercise 3.4.4 of Analysis I we then have $X^{-1}\left({\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{R}_i∕S\right)=X^{-1}\left({\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{R}_i\right)∕X^{-1}\left(S\right)$. With both sets being measurable in $F$, their set difference must be measurable as well.
• Let $S_1,S_2,\dots \subseteq {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{R}_i$ be measurable sets. By Exercise 3.4.3 of Analysis I we then have $X^{-1}\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{S}_i\right)\in F={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{X}^{-1}\left(S_i\right)\in F$ with the latter contained in $F$ by the properties of $\sigma$-algebra.