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Let $(X,\mathcal F)$ be a measure space, then for each $x \in X$ does there always exists a smallest measurable set containing $x$?

If $X$ is countable or $\mathcal F$ finite, then this is true, as then the set $$ \bigcap_{\substack{E \in \mathcal F \\ x \in E}} E $$ could be rewritten as an at most countable intersection (i.e. selecting just a countable subset of the measurable sets containing $x$). But what in the general case? If I look for example at the Borel-$\sigma$-algebra over $\mathbb R$, then each singleton set $\{x\}$ is measurable, and trivially the smallest set containing $x$, so there comes no example to my mind where such a set is not uniquely specified?

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    $\begingroup$ You're saying if $X$ is countable, then the set of all measurable sets containing a singleton is always countable? Why is that? ${}\qquad{}$ $\endgroup$ – Michael Hardy May 18 '15 at 18:47
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    $\begingroup$ No, I am saying that the intersection I have written above is equal to an at most countable intersection (not that the resulting set is itself countable or that the measurable sets which contain $x$ itself form a countable family of sets). For a proof, where this set intersection is rewritten such that the sets over which is intersected are selected in accordance with elements from $X$ (and thereby at most countable if $X$ is), see: math.stackexchange.com/questions/931744/… $\endgroup$ – StefanH May 18 '15 at 18:54
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    $\begingroup$ Okay, my wording was a little bit misleading, I edited my post, hopefully it is clear now? I do not mean that the set of all $\{ E \}$ with $x \in E$ is countable, I mean that you can select some countable family of sets whose intersection is equal to the intersection of all the $E$'s. $\endgroup$ – StefanH May 18 '15 at 18:58
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The answer is no, and here is one example: Let $X=\mathbb{R}$, and define a set $E \subseteq \mathbb{R}$ to be measurable if either

  • $0 \notin E$ and $E$ is countable, or
  • $0 \in E$ and the complement $E^c$ is countable.

Then it is easy to see that there is no smallest measurable set containing $x=0$.

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Like Lukas Geyer has shown, in general no.

But if you can find a finite measure $\mu$ on $(X,\mathcal{F})$ such that $\mu(A) = 0 \implies A = \emptyset$ for all $A \in \mathcal{F}$ (i.e. the empty set is the only null set), then it is true:

Let $x \in X$ and $\alpha = \inf\{\mu(A)\mid x \in A \in \mathcal{F}\}$. Then, since $\mu$ is finite, $\alpha < \infty$ and hence there is a sequence $x \in A_n \in \mathcal{F}$ such that $\lim_{n \to \infty} \mu(A_n) = 0$. Let $$B = \bigcap_{n=1}^\infty A_n.$$ Then $x \in B \in \mathcal{F}$ and $\mu(B) = \alpha$. Obviously $$E_x := \bigcap_{x \in A \in \mathcal{F}}A \subseteq B.$$ Suppose that $B \not \subseteq E_x$. Then there is some $y \in X$ and $x \in C \in \mathcal{F}$ such that $y \in A_n$ for all $n \in \mathbb{N}$ and $y \not \in C$. Then $$\alpha = \mu(B) = \underbrace{\mu(B\setminus C)}_{>0} + \underbrace{\mu(B \cap C)}_{= \alpha}$$ where $\mu(B\setminus C) > 0$ by $y \in B\setminus C$ and the property of $\mu$. Contradiction.

In fact it is easy to see that it is enough that $\alpha < \infty$, i.e. that the measure $\mu$ satisfies that for each $x \in X$ there is some $x \in A \in\mathcal{F}$ such that $\mu(A) < \infty$. In particular, a $\sigma$-finite measure $\mu$ equivalent to the counting measure is enough.

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