Let $Q$ be a probability measure on $(\mathbb{R},\beta(\mathbb{R}))$ such that for all $B\in\beta(\mathbb{R})$, it holds that $Q(B)\in\{0,1\}$. Show that there exists an $a\in\mathbb{R}$ such that $Q=\delta_a$.

where $\beta$ refers to Borel $\sigma$ algebra

I think it's something to do with bisection of intervals, but I don't know where to go with that.



Prove that:

  • It exists $n_0 \in \mathbb N$ such that $Q([n_0, n_0+1))=1$.
  • This $n_0$ is unique.
  • Build by induction a sequence of decreasing intervals $\{I_n\}$ included in $[n_0,n_0+1)$ of lengths $1/2^n$ such that $Q(I_n)=1$.
  • The intersection of those interval is equal to a point $a$.
  • Conclude that $Q=\delta_a$.
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  • $\begingroup$ I don't think uniqueness of $n_0$ is required. It will automatically follow from $Q(\{a\})=1$, which is already sufficient for $Q=\delta_a$. $\endgroup$ – Vercassivelaunos Aug 24 at 20:35
  • $\begingroup$ While this is true, proving straightforward $n_0$ uniqueness eases the construction of the intervals $I_n$. $\endgroup$ – mathcounterexamples.net Aug 24 at 20:41
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    $\begingroup$ I see! One last thing though: It should probably be mentioned that $I_n$ should all be closed intervals, otherwise their intersection might be empty. Just because the initial interval was constructed as half open, which makes it tempting to choose the others as half open as well. $\endgroup$ – Vercassivelaunos Aug 24 at 20:49
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    $\begingroup$ @Vercassivelaunos: the intersection cannot be empty, because it has measure 1. But the proof should mention it. $\endgroup$ – André Caldas Aug 24 at 22:04
  • $\begingroup$ It is nice to notice that the intervals do not need to be disjoint. As long as you can cover the space with a countable number of arbitrarily small sets. $\endgroup$ – André Caldas Aug 24 at 22:06

Bisection of intervals is the right idea. First show that there exists a half open interval $\tilde I$ with $Q(\tilde I)=1$. This is true because $\mathbb R$ can be written as a countable disjoint union of such intervals, and $Q(\mathbb R)=1$, so due to $\sigma$-additivity, they can't all have measure $0$. Then define $I$ as the closure of $\tilde I$. Now construct a sequence of intervals $I_n$ by bisecting $I$ and choosing the half with measure $1$, and repeating iteratively. Always choose the intervals to be closed, so we can use the nested intervals theorem in the end. Now $I_n$ is a sequence of closed nested intervals whose length converges to $0$. Since $\mathbb R$ is complete, their intersection is a singleton set (nested intervals theorem), and due to $Q$'s continuity from above, the measure of that singleton is $1$. The one element inside it is $a$.

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