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Let $X$ be a set, $F$ a $\sigma$-field of subsets of $X$, and $\mu$ a measure on $F$. A map $F=(f_1,\ldots,f_n)$ of $X$ into $\mathbb{R}^n$ is said to be measurable if each of the coordinate functions, $f_i$, is measurable. Show that $f$ is measurable if and only if $f^{-1}(B)\in F$ for every Borel set $B\subseteq \mathbb{R}^n$.

The function $f_i$ being measurable means that $f_i^{-1}(B)\in F$ for every Borel set $B\in\mathbb{R}$. Therefore, we must show that this is true for every $i=1,\ldots,n$ if and only if $f^{-1}(B)\in F$ for every Borel set $B\subseteq \mathbb{R}^n$.

Borel sets are the smallest $\sigma$-ring containing the open sets. I can't see how to relate Borel sets in $\mathbb{R}$ and Borel sets in $\mathbb{R}^n$.

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    $\begingroup$ The $\sigma$-algebra of the Borel sets on $\mathbb R^n$ is generated by sets of the form $X_1\times\dots \times X_n$, where each $X_i\subset \mathbb R$ is a Borel subset. To see it notice that it is generated already by products of open intervals, since these products generate the topology on $\mathbb R^n$. $\endgroup$ – user8268 Sep 18 '13 at 20:19
  • $\begingroup$ Show that for any set $X$ and $\sigma$-algebra $\mathcal{A}$ on $X$, and every function $g\colon X \to Y$, the family $\mathcal{B} = \{ B \subset Y : g^{-1}(B) \in \mathcal{A}\}$ is a $\sigma$-algebra on $Y$. $\endgroup$ – Daniel Fischer Sep 18 '13 at 20:20
  • $\begingroup$ @DanielFischer I can show that, but I can't see how to apply it here. $\endgroup$ – PJ Miller Sep 18 '13 at 20:31
  • $\begingroup$ If all of the $f_i$ are measurable, then that $\sigma$-algebra contains all sets of the form $\{ x_i > c\}$. Hence all complements of such sets, countable intersections and unions of such. Thus it contains all products of open intervals with rational endpoints. These are a countable basis of the topology of $\mathbb{R}^n$, hence the $\sigma$-algebra contains all open sets. Hence all Borel sets. Conversely, if $\mathcal{B}$ contains all Borel sets, it contains in particular all sets of the form $\{x_i > c\}$, which means that $f_i$ is measurable. $\endgroup$ – Daniel Fischer Sep 18 '13 at 20:37
  • $\begingroup$ @user8268 Could you please elaborate on your idea. I am struggling to see how you could write your idea as a solution to the original problem. $\endgroup$ – Permian Jun 16 '14 at 17:28

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