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Prove that Borel sigma-field on $\mathbb{R}^d$ is the smallest sigma-field that makes all continuous functions $f:\mathbb{R}^d \to \mathbb{R}$ measurable.

How do I go about proving this? Thanks!

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It is only a matter of definitions. Recall that a continuous function is one for which the preimage of an open set is an open set and that the Borel sigma-field is the smallest sigma-field containing the open sets. –  Giuseppe Negro Sep 1 '12 at 8:09
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. Titles should be informative. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative to be rude when asking for help; please consider rewriting your post. –  Michael Greinecker Sep 1 '12 at 10:23
    
To show that the smallest $\sigma$-algebra making all continuous functions measurable is at least as large as the Borel $\sigma$-algebra, you can use that $\mathbb{R}^n$ is perfectly normal. –  Michael Greinecker Sep 1 '12 at 10:34
    
@MichaelGreinecker Sorry but I am not sure I subscribe to your suggestion. (+1 for your first comment, though.) –  Did Sep 1 '12 at 10:37
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@MichaelGreinecker The problem is the discrepancy between the level of sophistication of the (perfectly valid) approach you suggest and (what one can guess about) the OP's mathematical maturity, knowing that a fairly direct way indeed exists. –  Did Sep 1 '12 at 10:42

1 Answer 1

Let $F\subseteq\mathbb{R}^n$ be a nonempty closed set. We let $\delta_F:\mathbb{R}^n\to\mathbb{R}$ be the distance from $F$, that is, $$\delta_F(x)=\inf\big\{\|y-x\|:y\in F\big\}$$ for all $x\in\mathbb{R}^n$. You can show that $\delta_F(x)=0$ if and only if $x\in F$, so $F=\delta_F^{-1}\big(\{0\}\big)$, and $\{0\}$ is a Borel subset of $\mathbb{R}$. So all closed sets are in the smallest $\sigma$-field that makes all continuous functions measurable. The rest should be manageable.

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