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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.

Manolis D
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    (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. – Fedor Petrov Jan 21 '21 at 15:33
  • @Fedor: (i) By this logic, I guess I can basically take any measure whose support lies outside $A$. (ii) Ah, right. I keep thinking about the infinite-dimensional $X$ so don't see the obvious. Thanks for pointing that out. – Manolis D Jan 21 '21 at 16:26
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    (i) sure you can, this is rather tautological. But sometimes a measure with greater supports (intersecting $A$) may also work. – Fedor Petrov Jan 21 '21 at 16:28
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    Perhaps a more interesting question is: is there a Borel probability measure on $\mathbb{R}$ with respect to which every meager set is null? – Nik Weaver Jan 21 '21 at 16:50
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    @NikWeaver: This was answered here, the answer is no: https://mathoverflow.net/questions/342798/is-there-a-measure-on-0-1-that-is-0-on-meagre-sets-and-1-on-co-meagre-sets?rq=1 – Christian Remling Jan 21 '21 at 16:57
  • @ChristianRemling ah, nice! – Nik Weaver Jan 21 '21 at 17:41
  • (i) You did not require completeness, you could easily have $X$ first category in itself. For example, the subspace of normed space $l^2$ consisting of sequences that are zero except for a finite number of coordinates. A normed space is certainly locally convex and metrizable. – Gerald Edgar Jan 21 '21 at 17:58
  • (ii) I think this is likely false as well. Can we prove (or disprove) this for the classical Wiener measure on $C[0,1]$, for example? – Gerald Edgar Jan 21 '21 at 18:06
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    @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. – Nate Eldredge Jan 21 '21 at 18:10

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