Exercise 1.2

Question

Prove an analogue of Theorem $1.8$ for $n$ functions.

Amit Awasthi · Accepted Answer

\newtheorem*{claim}{Claim}

\begin{claim}
      Let $u_1,\ldots,u_n$ be real measurable functions on a measurable space
      $X$, let $\Phi$ be a continuous mapping of $\mathbf{R}^n$ into a
      topological space $Y$, and define
      \begin{equation*}
        h(x) = \Phi(u_1(x),\ldots,u_n(x))
      \end{equation*}
      for $x \in X$. Then $h \colon X \to Y$ is measurable.
    \end{claim}
    \begin{proof}
      Put $f(x) = (u_1(x),\ldots,u_n(x))$. Then $f$ maps $X$ into
      $\mathbf{R}^n$. Since $h = \Phi \circ f$, by Theorem $1.7(b)$, it suffices
      to show $f$ is measurable.
      \par If $R$ is any cartesian product of open intervals $I_i$ in
      $\mathbf{R}^n$, then
      $$ f^{-1}(R) = \bigcap_{i=1}^n u_i^{-1}(I_i) $$
      which is measurable. Every open set $V \subset mathbf{R}^n$ is the countable
      union of such rectangles $R_j$, and so
      \begin{equation*}
        f^{-1}(V) = f^{-1}\!\left( \bigcup_{j=1}^\infty R_j \right) =
        \bigcup_{j=1}^\infty f^{-1}(R_j)
      \end{equation*}
      is measurable.
    \end{proof}

Exercise 1.2

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Exercise 1.2

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