Is every element contained in a smallest measurable set?

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?

• You're saying if $X$ is countable, then the set of all measurable sets containing a singleton is always countable? Why is that? ${}\qquad{}$ – Michael Hardy May 18 '15 at 18:47
• 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/… – StefanH May 18 '15 at 18:54
• 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. – StefanH May 18 '15 at 18:58

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 $x=0$.

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.