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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.2012-09-01
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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](http://en.wikipedia.org/wiki/Perfectly_normal_space).2012-09-01
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    @MichaelGreinecker Sorry but I am not sure I subscribe to your suggestion. (+1 for your first comment, though.)2012-09-01
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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.2012-09-01
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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.2012-09-01
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    @did Point taken.2012-09-01

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