Let $(X, \mathcal B, \mu)$ be a "good" measure space, e.g. $\mu$ is a positive Radon measure on a locally compact topological space $X$ with Borel $\sigma$-algebra $\mathcal B$. Let $A\subset X$ such that every measurable subset of $A$ has zero measure. Is it true that there is a zero measure set $B$ such that $A\subset B$?
[Math] Is a subset that contains no positive measurable subsets contained in a null measurable set
measure-theory
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Tautological answer: the cases where you know the measure of compact sets (as in section 3 of chapter IX of Bourbaki), but not the class of all measurable sets (which you then can define using the above definition applied to the indicator function of the set). So the next question is: should one take the topology (or even only the compact subsets), and not the class of all measurable set, as starting point? Answer: perhaps. To understand the answer, one can look at the review of Bourbaki's integration by G.A.Edgar, pag. 82-84 of Mathematical intelligencer, 1981 (I have no month in my copy of that review). It is worth reading; among other things one reads:
There are valid motives for doing measure theory using Radon measures instead of abstract measures. An important one is the problem of disintegration of measures (or of "regular conditional probabilities"). [...] Another problem with abstract measures concerns the connection between point-maps and set-maps. [...] A third shortcoming of abstract measures is called the "image-measure catastrophe" by L.Schwartz [in his book "Radon measures on arbitrary topological spaces and cylindrical measures"]. [...] On the other hand, there are also many reasons to use abstract measures instead of Radon measures.
For these reasons, he cites the preceding reviews of Bourbaki's integration by Halmos and by Munroe in the bull. AMS: 59 (1953) and 65 (1958); you can find them easily on line. He cites another reason, the "barycenter catastrophe".
Even A.Weil himself was not completely satisfied with Bourbaki's integration (but for reasons different from the requester); the see Weil's comments in the third volume of his collected papers, where he refers to a compact system of sets as starting point and Schwartz's book for some more details. I only remark that Weil's starting point would provide a connection with one concept categorical topologist studied around 1970: compactly generated (weak) Hausdorff spaces as convenient setting for topology (convenient meaning that the exponential law is available)
No. Counterexamples include $X = \{0,1\}^\kappa$ or $[0,1]^\kappa$ for any $\aleph_0 < \kappa \le \mathfrak{c}$. These spaces are compact Hausdorff (Tikhonov's theorem) and are separable (Hewitt-Marczewski-Pondiczery).
(This special case of H-M-P can also be proved directly. For example, with $X =[0,1]^\kappa$, identify $\kappa$ with a subset of $[0,1]$, so that $X$ is a space of functions from $[0,1]$ to itself; then it is easy to show that the set of polynomials with rational coefficients is dense in $X$.)
Now the uncountable successor ordinal $\omega_1 + 1$, with its order topology, can be embedded in $X = \{0,1\}^\kappa$ or $X = [0,1]^\kappa$. See Fremlin, Measure Theory, 434K(d) (thanks to Robert Furber for the reference). One can also follow a construction similar to the Stone-Čech compactification and embed $\omega_1 + 1$ into $[0,1]^{C(\omega_1+1, [0,1])}$, noting that $|C(\omega_1 + 1, [0,1])| = \mathfrak{c}$. Either way, call the embedding $\phi$.
Let $\nu$ be the famous Dieudonné measure on $\omega_1 + 1$, where $\nu(B) = 1$ if $B$ contains a closed unbounded subset of $\omega_1$ and $\nu(B) = 0$ otherwise. (See Fremlin 411Q and 4A3J for details.) Then the pushforward $\mu = \nu \circ \phi^{-1}$ is a finite Borel measure on $X$ which is not regular.
Specifically, since $\phi$ is an embedding and $\omega_1$ is open in $\omega_1 + 1$, then the image $A = \phi(\omega_1)$ is a relatively open subset of the compact set $\phi(\omega_1 + 1)$, so that $A$ is the intersection of an open set and a closed set in $X$, and in particular is Borel. Moreover $\mu(A) = \nu(\omega_1) = 1$. But if $K$ is any compact subset of $A$, then $\phi^{-1}(K)$ is a compact subset of $\omega_1$, hence $\mu(K) = \nu(\phi^{-1}(K)) = 0$. Thus inner regularity fails.
This example is also Exercise 7.14.130 of Bogachev's Measure Theory.
Best Answer
You are asking whether every set with inner measure $0$ has measure $0$ with respect to the completion measure. The Lebesgue measure, for example, does not have this property, since the usual Vitali set is non-measurable, but has inner measure $0$.
I claim that a ($\sigma$-finite) measure space has your property if and only if every set is measurable with respect to its completion measure.
For the forward direction, suppose that $A$ is any subset of the space $X$. Let $a$ be the inner measure of $A$, the supremum of $\mu(A_0)$ among all measurable $A_0\subset A$ (and we may assume wlog this is finite). By taking a union, it follows that the inner measure is realized, so that there is some measurable $A_0\subset A$ with $\mu(A_0)=a$. It follows that $A-A_0$ has inner measure $0$. By the hypothesis, it follows that $A-A_0\subset B$ for some measurable set $B$ with $\mu(B)=0$, and so $A-A_0$ has measure $0$ with respect to the completion. Consequently, $A$ differs from the measurable set $A_0$ on a completion-measure zero set $A-A_0$, and hence is measurable with respect to the completion measure.
Conversely, suppose that every set is measurable with respect to the completion of $\mu$. Suppose that $A$ has inner measure $0$. By assumption, there is some measurable set $A_0$ such that the symmetric difference $A\triangle A_0\subset A_1$ for some measurable $A_1$ with $\mu(A_1)=0$. It follows that $A_0-A_1$ is a measurable subset of $A$, and hence measure $0$, and so $A\subset B=A_0\cup A_1=(A_0-A_1)\cup A_1$ shows that $A$ is contained in a measure $0$ set $B$, as desired.
In particular, a complete measure has the property if and only if it measures every set.