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Let $\mu$ be a non-atomic probability measure defined on $\mathbb R$. Let $I$ simply be an interval of length $\beta$, i.e., $ I = [i \beta,(i+1)\beta]$ for some $i \in \mathbb{Z}$, and $B$ any Borel set. I have the following fact given to me, which I was not able to refute yet and might well be correct:

$\mu(I \cap B) \rightarrow \mu(I) \dfrac{\mathrm{Len}(I \cap B)}{\mathrm{Len}(I)}$ as $\beta\downarrow 0$.

Here $\mathrm{Len}$ denotes the 1-D volume or the exterior measure.

This is a bit hard to imagine. The intuition (which may fail most of the times at measure theory) says that this is roughly dividing the probability assigned to interval $I$ by its length so that we have a 'unit/uniform' measure now. Then we assign this unit measure to the length of $I \cap B$ simply by multiplication. How can we prove this result, if this is true? If not, what are the minimum number of assumptions we need such that this is true?

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    $\begingroup$ How are you defining $Len(I \cap B)$? $\endgroup$ May 19, 2021 at 14:40
  • $\begingroup$ @user6247850 I did not do that yet, I was hoping for the 1-D volume (exterior measure). $\endgroup$ May 19, 2021 at 15:03
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    $\begingroup$ If $\mu$ is non-atomic then as $\beta \to 0$, $\mu(I \cap B) \to 0$. So in some sense you want to claim that this is the "leading order term" in the asymptotic expansion as $\beta \to 0$. (Or refute that, too.) $\endgroup$
    – Ian
    May 19, 2021 at 15:31
  • $\begingroup$ So it seems like perhaps you want $\sim$ in place of $\to$ in the statement. I am also not quite sure of the role of $i$; is it fixed once and for all, or what? $\endgroup$ May 24, 2021 at 4:17
  • $\begingroup$ @NateEldredge thanks for your comment. $i$ is defined such that $I$ is an interval of length $\beta$. I can change my wording there to be more specific. $\endgroup$ May 24, 2021 at 15:31

2 Answers 2

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I interpret your question as follows.

Let $\mu$ be a non-atomic probability measure on $\mathbb{R}$. Let $B$ be a Borel subset of $\mathbb{R}$. Then, $$\lim_{\beta \downarrow 0} \frac{\mu([0,\beta]\cap B)/\mu([0,\beta])}{|[0,\beta]\cap B|/|[0,\beta]|} = 1,$$ where $|\cdot|$ denotes Lebesgue measure.

This claim is definitely false. One can take $\mu = \sum_{k=1}^\infty 2^{-k}\text{Unif}([2^{-2^k-1},2^{-2^k}])$, where $\text{Unif}(I)$ denotes the uniform probability measure over $I$, and $B = \cup_{k=1}^\infty [2^{-2^k-2},2^{-2^k}]$.

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I think it may not be correct for all cases. Consider a positive decreasing sequence $\beta_n \to 0$ for $n \to \infty$. We have that $\textrm{Len}(I_n)=\beta_n$. Let's consider $I_n=[0,\beta_n]$ (i.e. $i=0$). We have that $I_1 \supset I_2 \supset I_3 \supset ...$ and $I_\infty=\cap_{n \in \mathbb{N}}I_n=\{0\}$. So $\lim_{n \to \infty}\mu(I_n)=\mu(\{0\})$ and $\lim_{n \to \infty}\mu(I_n \cap B)=\mu(B \cap \{0\})$. Let $B$ be $(-\varepsilon,0]$ for some $\varepsilon >0$. We have that $\textrm{Len}(I_n \cap B)=\textrm{Len}(\{0\})=0,\forall n \in \mathbb{N}$. So $$\lim_{n \to \infty}\frac{\textrm{Len}(I_n \cap B)}{\textrm{Len}(I_n)}\mu(I_n)=\lim_{n \to \infty}0 = 0$$ which is not necessarily equal to $\mu(B \cap \{0\})=\mu(\{0\})$.

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  • $\begingroup$ Thanks for your answer! However, isn't probability measure on a singleton always $0$? $\endgroup$ May 19, 2021 at 15:02
  • $\begingroup$ A probability measure is a measure on a set $X$ such that $\mu(X)=1$, nothing else $\endgroup$
    – Snoop
    May 19, 2021 at 15:06
  • $\begingroup$ That is a good point, thank you. Which additional assumptions are needed so that the statement above is correct? $\endgroup$ May 19, 2021 at 15:12
  • $\begingroup$ I think the measure should be non-atomic, am I right? $\endgroup$ May 19, 2021 at 15:22
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    $\begingroup$ Added that, thanks so much! $\endgroup$ May 19, 2021 at 15:27

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