- \(\star\): Show that cylinder functions are measurable, i.e. if \(f\) is measurable on \(\RR^s\), then \(F(x, y) \definedas f(x)\) is measurable on \(\RR^s\cross \RR^t\) for any \(t\).
Solution
1. The claim holds when \(f = \chi_E\) is the indicator of a measurable set \(E \subseteq \RR^s\).
1.1. \(F = \chi_{E \cross \RR^t}\).
\(F(x,y) = 1\) if and only if \(x \in E\), that is, if and only if \((x,y) \in E \cross \RR^t\).
1.2. \(E \cross \RR^t\) is measurable.
Write \(E = V \setminus Z\) with \(V\) a \(G_\delta\) set and \(Z\) a Lebesgue null set. Then \(E \cross \RR^t = (V \cross \RR^t) \setminus (Z \cross \RR^t)\). The set \(V \cross \RR^t = \pi_s^{-1}(V)\) is Borel, because the projection \(\pi_s: \RR^s \cross \RR^t \to \RR^s\) is continuous. The set \(Z \cross \RR^t\) is null: \(Z \cross [-k,k]^t\) is covered by products of cubes of total measure at most \((2k)^t\eps\) for every \(\eps > 0\), and \(Z \cross \RR^t\) is the countable union over \(k\) of these sets. So \(E \cross \RR^t\) is Lebesgue measurable.
1.3. Q.E.D.
Steps 1 and 2 show that \(F\) is the indicator of a measurable set.
2. The claim holds for simple functions \(f = \sum_{i=1}^k c_i \chi_{E_i}\).
\(F(x,y) = \sum_i c_i \chi_{E_i}(x)\) is a finite linear combination of functions that are measurable by step 1.
3. The claim holds for measurable \(f \geq 0\).
There are simple functions \(0 \leq s_n \nearrow f\) pointwise. By step 2, \(F_n(x,y) \coloneqq s_n(x)\) is measurable, and \(F_n(x,y) \nearrow f(x) = F(x,y)\) for each \((x,y)\). A pointwise limit of measurable functions is measurable.
4. Q.E.D.
For general measurable \(f\), write \(f = f^+ - f^-\) and apply step 3 to \(f^+\) and \(f^-\).