Let $f$ be a complex-valued continuous function on Wiener space such that $|f|$ is measurable. Is $f$ then measurable, too? I am looking for a proof or a counterexample.
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2$\begingroup$ If $f$ is continuous then it's certainly measurable, no further assumptions needed, assuming you mean measurable with respect to the Borel $\sigma$-algebra. Did you mean to write something else? $\endgroup$– Nate EldredgeCommented Jul 31, 2018 at 16:30
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$\begingroup$ @NateEldredge; Where can I find a detailed argument? I mean the sigma algebra with respect to which the Gaussian measure is defined. $\endgroup$– Arnold NeumaierCommented Jul 31, 2018 at 19:03
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1$\begingroup$ Classical Wiener measure, on the space of continuous functions, is defined on the Borel $\sigma$-algebra (or its completion). And every continuous function is measurable with respect to the Borel $\sigma$-algebra. This is completely trivial and there really are no details: under a continuous function, the preimage of every open set is open and hence Borel, so the function is measurable. $\endgroup$– Nate EldredgeCommented Jul 31, 2018 at 19:07
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