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 01 '12 at 08:09
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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 01 '12 at 10:34
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@MichaelGreinecker Sorry but I am not sure I subscribe to your suggestion. (+1 for your first comment, though.) – Did Sep 01 '12 at 10:37
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@did What is the problem? It allows you to write every closed set as the preimage of a singleton under a continuous function. Of course, that can be done in a fairly direct way too. – Michael Greinecker Sep 01 '12 at 10:40
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2@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 01 '12 at 10:42
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@did Point taken. – Michael Greinecker Sep 01 '12 at 10:43
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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.
Michael Greinecker
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