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It is well known that meagre sets (in topology theory) and sets of measure zero (in measure theory) are generally not the same things ([O], see also a related question on MO). A set that is small in one sense may be large in the other sense. But surely, from the names suggested, should the opposite be true at least for some common regular measures? In particular, let $X$ be a locally convex metrizable space, I would like to know

(i) If $A$ is a set of the first category of $X$, does there exist a (probability) regular measure $\mu$ on $X$ such that $\mu(A) = 0$? 

(ii) Would the Gaussian measure on $X$ be a positive answer for (i)? 

Reference: [O] Oxtoby, J. C., Measure and category. A survey of the analogies between topological and measure spaces, Graduate Texts in Mathematics. 2. New York-Heidelberg-Berlin: Springer- Verlag. VIII, 95 p. (1971). ZBL0217.09201.

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    $\begingroup$ (i) say, a $\delta$-measure concentrated in a point outside $A$. (ii) if $X=\mathbb{R}$, a Gaussian measure is equivalent to Lebesgue measure and it does not always work. $\endgroup$
    – Fedor Petrov
    Commented Jan 21, 2021 at 15:33
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    $\begingroup$ (i) sure you can, this is rather tautological. But sometimes a measure with greater supports (intersecting $A$) may also work. $\endgroup$
    – Fedor Petrov
    Commented Jan 21, 2021 at 16:28
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    $\begingroup$ Perhaps a more interesting question is: is there a Borel probability measure on $\mathbb{R}$ with respect to which every meager set is null? $\endgroup$
    – Nik Weaver
    Commented Jan 21, 2021 at 16:50
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    $\begingroup$ @NikWeaver: This was answered here, the answer is no: mathoverflow.net/questions/342798/… $\endgroup$
    – Christian Remling
    Commented Jan 21, 2021 at 16:57
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    $\begingroup$ @GeraldEdgar: As noted in the post linked by Christian Remling, for every Borel probability measure on a metric space, there is a meager set of full measure. With Wiener measure, a nice example is to take the Holder space $A = C^{0,\alpha}[0,1] \subset C[0,1]$ for $\alpha < 1/2$. It's a standard fact that Brownian motion is a.s. $\alpha$-Holder-continuous for $\alpha < 1/2$, so $\mu(A) = 1$, but $A$ is easily seen to be $\sigma$-compact (by Arzela-Ascoli) and therefore meager. $\endgroup$
    – Nate Eldredge
    Commented Jan 21, 2021 at 18:10

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