Ex4-3

=== Problem 4.3 ===
Theorem 4.3 can be used to define measurability for vector-valued (e.g., complex-valued) functions. Suppose, for example, that f and g are real-valued and finite in \mathbb{R}^n, and let F(x)=\left(f(x),g(x)\right). Then F is said to be measurable if F^{-1}(G) is measurable for every open G\subset\mathbb{R}^2. Prove that F is measurable if and only if both f and g are measurable in \mathbb{R}^2. 
=== Solution. ===
Suppose that F is measurable, then \left\{f>a\right\}=F^{-1}\left(\left(a,\infty\right)\times \mathbb{R}\right) and \left\{g>a\right\}=F^{-1}\left(\mathbb{R}\times\left(a,\infty\right)\right) are measurable for all a\in\mathbb{R} since both \left(a,\infty\right)\times\mathbb{R}\subset \mathbb{R}^2 and \mathbb{R}\times\left(a,\infty\right)\subset \mathbb{R}^2 are open subsets of \mathbb{R}^2. This shows that both f and g are measurable in \mathbb{R}^2. Conversely, if both f and g are measurable in \mathbb{R}^2, then for every open set G\subset\mathbb{R}^2 that can be written as the union of closed rectangles whose corners have rational coordinates, we haveF^{-1}(G)=F^{-1}\left(\bigcup_k \left[a_k,b_k\right]\times \left[c_k,d_k\right]\right)=\bigcup_k F^{-1}\left(\left[a_k,b_k\right]\times \left[c_k,d_k\right]\right)=\bigcup_k\left[\left\{a_k\leq f\leq b_k\right\}\cap \left\{c_k\leq g\leq d_k\right\}\right]which is a countable union of measurable sets, thus F is a measurable set.\blacksquare