Exercise 1.7.21 (Associativity of product of measure spaces)

We must demonstrate that

We use the usual trick of demonstrating that the set on the left-hand side is both the subset and the superset of the set on the right-hand side.

• The trick is to rewrite the theorem assertion as:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->D\in B_X\times B_Y\forall <!--\; FUNCTION\; APPLICATION\; -->C\in B_Z:D\times C\in B_X\times \left(B_Y\times B_Z\right).$$

  This formulation gives us a new attack angle; namely, denoting

  $$P(D):=\text{for all }C\in B_Z:D\times C\in B_X\times \left(B_Y\times B_Z\right)$$

  we can restrict ourselves only to the sets $D$ contained in the generating set $B_0=\{E\times F\mid E\in B_X,F\in B_Y\}$ instead of the whole $\sigma$-algebra $B_X\times B_Y$ by employing the famous measurable induction principle. We verify its assumptions.

  (i)

  For any $C\in B_Z$ we have $\varnothing \times C=\varnothing \in B_X\times \left(B_Y\times B_Z\right)$ as the latter set is a $\sigma$-algebra and must therefore contain the empty set.

  (ii)

  Suppose that $(E\times F)\times C\in B_X\times \left(B_Y\times B_Z\right)$ for any $C\in B_Z$. Fix $C$. Since the product set is a $\sigma$-algebra, it must also contain the complement, which by inductive version of Exercise 3.5.5 of Analysis I is equal to $$\begin{array}{llll}\hfill X\times Y\times Z\setminus \left(E\times F\times C\right)& =\left[\left((X\times Y)\setminus (E\times F)\right)\times Z\right]\bigcup \left[(X\times Y)\times (Z\setminus C)\right]& \hfill & \\ \hfill & =\left[\left((X\setminus E)\times Y\bigcup X\times (Y\setminus F)\right)\times Z\right]\bigcup \left[(X\times Y)\times (Z\setminus C)\right]& \hfill & \\ \hfill & =\left[(X\setminus E)\times Y\times Z\right]\bigcup \left[X\times (Y\setminus F)\times Z\right]\bigcup \left[(X\times Y)\times (Z\setminus C)\right]& \hfill & \\ \hfill & =E^C\times Y\times Z\bigcup X\times F^C\times Z\bigcup X\times Y\times C^C& \hfill & \end{array}$$

  Furthermore, $B_X\times \left(B_Y\times B_Z\right)$ is generated by $\{E\times G\mid E\in B_X,G\in B_Y\times B_Z\}$; thus, both the sets $X\times (Y\times C^C)$ and $X\times (Y\times C)$ are contained in $B_X\times \left(B_Y\times B_Z\right)$. We conclude, by associativity of the Cartesian product (Example 3.5.9 of Analysis I), that

  $${(E\times F)}^C\times C=(E\times F\times C)\setminus X\times Y\times C^C\bigcup X\times Y\times C\in B_X\times \left(B_Y\times B_Z\right).$$

  (iii)

  Let $E_1\times F_1,E_2\times F_2,\dots \subseteq X\times Y$ be such that for all $C\in B_Z$ we have $(E_n\times F_n)\times C\in B_X\times \left(B_Y\times B_Z\right)$. Then, by the analogous property of the Cartesian product, the countable union can be expressed as $$\left[\bigcup _{n=1}^{\infty }{E}_n\times F_n\right]\times C=\bigcup _{n=1}^{\infty }\left[\left(E_n\times F_n\right)\times C\right]\in B_X\times \left(B_Y\times B_Z\right)$$

  as the latter is a $\sigma$-algebra.

  Finally, we verify whether this property is true for all $E\times F\in B_X\times B_Y$, i.e,

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->D\in \{E\times F\mid E\in B_X,F\in B_Y\}\forall <!--\; FUNCTION\; APPLICATION\; -->C\in B_Z:D\times C\in B_X\times \left(B_Y\times B_Z\right).$$

  The above is obviously true, since any set of the form $(E\times F)\times C$ for $E\in B_X,F\in B_Y,C\in B_Z$ can be expressed as $E\times (F\times C)$ by associativity of the Cartesian product, and it is therefore contained in $\sigma$-algebra $B_X\times \left(B_Y\times B_Z\right)$ generated by the set $\{E\times G\mid E\in B_X,G\in B_Y\times B_Z\}$.

• The second direction follows completely symmetrically to the previous part, so we do not provide its proof here.