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Oxymoron

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__QUESTION 1:__"

*If [itex]f\,:\,X \rightarrow \mathbb{R}^n[/itex] is a function then if its coordinate functions are measurable then does this imply that f is measurable?*"

Suppose that we have a [itex]\sigma[/itex]-algebra, [itex]\mathcal{A}[/itex] defined on some set X. We say that a function [itex]f\,:\,X \rightarrow \mathbb{R}^n[/itex] is [itex]\mathcal{A}[/itex]-measurable if [itex]f^{-1}(B) \in \mathcal{A}[/itex] for every [itex]B \in \mathcal{B}^n[/itex], that is, if the pre-image of every Borel set is in the sigma-algebra.

My question is this: Would it be possible to prove that the function [itex]f\,:\,X\rightarrow\mathbb{R}^n[/itex] is measurable if and only if its coordinate functions [itex]f_i\,:\,X\rightarrow\mathbb{R}[/itex] are measurable? (Where the coordinate functions are characterised by [itex]f(x) = (f_1(x),\dots,f_n(x))[/itex] for all [itex]x \in X[/itex]).

I mean, is f measurable if and only if all of its coordinate functions are measurable?

I know that if one regards the standard topology on [itex]\mathbb{R}[/itex] then every open set is a countable union of open intervals. Hence the Borel sets are generated by the set of all open intervals [itex]\{(a,b)\,:\,a < b\}[/itex].

__QUESTION 2:__"

*If I have a function [itex]f\,:\,\mathbb{R}^n \rightarrow \mathbb{R}^m[/itex] which is continuous then does this mean that it is automatically Borel measurable?*"

Is it true that I should be basing my idea of "Borel measurable" on the following?:

*A function f is Borel measurable if the pre-image of any Borel set is in the sigma-algebra*

If so, then when any question asks to prove that a function is Borel measurable then do I simply have to prove that the pre-image of every Borel set is in the sigma-algebra?

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