# Probability measure of a small interval

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?

• How are you defining $Len(I \cap B)$? May 19, 2021 at 14:40
• @user6247850 I did not do that yet, I was hoping for the 1-D volume (exterior measure). May 19, 2021 at 15:03
• 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.)
– Ian
May 19, 2021 at 15:31
• 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? May 24, 2021 at 4:17
• @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. May 24, 2021 at 15:31

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 $$B = \cup_{k=1}^\infty [2^{-2^k-2},2^{-2^k}]$$.

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 $$\mu(B \cap \{0\})=\mu(\{0\})$$.

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