Probability measure of a small interval

Question

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?

Answer 1

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

Answer 2

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