Chapter 11: Problem 1

\(\operatorname{lf}(X, M)\) and \((Y, N)\) are measurable spaces, show that the projection maps \(\mathrm{pr}_{1}: X \times\) \(Y \rightarrow X\) and \(\mathrm{pr}_{2}: X \times Y \rightarrow Y\) defined by \(\mathrm{pr}_{1}(x, y)=x\) and \(\mathrm{pr}_{2}(x, y)=y\) are measurable functions,

Short Answer

Expert verified

Both projection maps \(\mathrm{pr}_{1}\) and \(\mathrm{pr}_{2}\) are measurable functions, as their pre-images map any measurable set to a measurable set in \(X \times Y\).

Step by step solution

Verify Definition of Measurable Function

A function is measurable if the pre-image of any measurable set under the function is also a measurable set. Let's prove \(\mathrm{pr}_{1}: X \times Y \rightarrow X\) and \(\mathrm{pr}_{2}: X \times Y \rightarrow Y\) are measurable by this definition.

Analyze Projection Map \(\mathrm{pr}_{1}\)

Take a set \(A \subseteq X\) that is \(\sigma\)-measurable. The pre-image of \(A\) under the function \(\mathrm{pr}_{1}\) is given by the set \(\mathrm{pr}_{1}^{-1}(A) = A \times Y\). As \(A\) is a measurable set and \(Y\) is a complete space, their Cartesian product is a measurable set in the product \(\sigma\)-algebra on \(X \times Y\). Therefore, for any measurable set \(A \subseteq X\), \(\mathrm{pr}_{1}^{-1}(A)\) is also measurable, so \(\mathrm{pr}_{1}\) is a measurable function.

Analyze Projection Map \(\mathrm{pr}_{2}\)

Similarly, for a set \(B \subseteq Y\) that is \(\sigma\)-measurable, the pre-image of \(B\) under the function \(\mathrm{pr}_{2}\) is given by the set \(\mathrm{pr}_{2}^{-1}(B) = X \times B\). As \(B\) is a measurable set and \(X\) is a complete space, their Cartesian product is a measurable set in the product \(\sigma\)-algebra on \(X \times Y\). Therefore, for any measurable set \(B \subseteq Y\), \(\mathrm{pr}_{2}^{-1}(B)\) is also measurable, so \(\mathrm{pr}_{2}\) is a measurable function.