$\mu$ is a Borel measure on $\mathbb R$.

Define $f:\mathbb{R\to \bar R},f(x)={\operatorname{lim inf}}_{r\to 0}{{\mu((x-r,x+r))}\over{r}}$.

Prove that $f$ is (extended) Borel measurable.

I think maybe we need to write $\{x:f(x)>a\}$ in the form of intersection and union of sets,but I don't know how to deal with the inferior limit(since it's not discrete).Thanks in advance!


2 Answers 2


1). To deal with $\liminf_{r\to 0}$, it can be shown that $\liminf_{r\to 0}$ = $\sup_{k}\inf_{r< 1 / k}$, where $k \in N$ and $r \in Q$. Futhermore inf's and sup's of Borel measurable functions are Borel measurable.

2). To show that there is some $l \in \mathbb R_+$ that for any $r \in (0, l)$, $g(x) = {\mu((x-r,x+r))}$ is Borel measurable we could use the fact, that $\{x:g(x)<a\}$ is open for any $a \in \mathbb R$ if and only if $g$ is upper semicontinuous. So our goal is to prove that $\limsup_{y \to x}g(y) \le g(x)$ (*).

3). Let's fix $x\in \mathbb R$. By the properties of measure we know that $\mu((x-r,x+r))$ is non-decreasing by $r$. So there is some $m\in[0,\infty]$ such that for all $r>m$, $\mu((x-r,x+r))=\infty$ and for all $r < m$, $\mu((x-r,x+r))<\infty$. If $m = 0$, inequation (*) is obvious.

4). Now, let's consider case of $m>0$ and fix $r<\frac {m} {2}$. Let's also choose $y_k \in \mathbb R$ such that $y_k \to x$. As $\limsup_{k \to \infty} 1_{(y_k-r, y_k+r)} \le 1_{(x-r,x+r)}$ and $\liminf_{k \to \infty}(1 - 1_{(y_k-r, y_k+r)}) \ge 1 - 1_{(x-r,x+r)}$, it follows that:

$\int_{(x-2r,x+2r)}(1 - 1_{(x-r,x+r)})d\mu \le \int_{(x-2r,x+2r)}\liminf_{k \to \infty}(1 - 1_{(y_k-r, y_k+r)})d\mu \le $ {by Fatou's lemma} $\le \liminf_{k \to \infty} \int_{(x-2r,x+2r)}(1 - 1_{(y_k-r, y_k+r)})d\mu$

which, by linearity of integrals gives us

$\mu((x-2r, x+2r)) - \mu((x-r, x+r)) \le \liminf_{k \to \infty}(\mu((x-2r, x+2r)) - \mu((y_k-r, y_k+r))) = \mu((x-2r, x+2r)) - \liminf_{k \to \infty}\mu((y_k-r, y_k+r))$

As $r < \frac {m} {2}$, it is true that $\mu((x-2r, x+2r)) < \infty$, so we can remove it from both parts of inequality, and after change of signs we get (*).

5). As we prooved that $g(x)$ is Borel measurable, our last step is to show that $g(x) \over r$ is Borel measurable, which is obvious, because constant function is measurable and measurable function divided by measurable function is also measurable.

  • 1
    $\begingroup$ Fix an r, we suppose that $μ((x−r,x+r)) < \infty$, but why is it also true that $μ((x−2r,x+2r))< \infty$? $\endgroup$
    – Maksim
    Mar 20 at 4:20
  • $\begingroup$ @Maksim I messed up with that. It shouldn't be true in general. I changed my anwer to fix that (fix is mostly in step 3). $\endgroup$
    – brovovar
    Mar 20 at 14:12
  • $\begingroup$ I think there is still some minor problem, since the choice of $m$ is dependent on $x$, we can not find some $l \in \mathbb{R}^+$ such that for any $r\in (0,l), g(x)=μ((x−r,x+r))$ is upper semicontinuous. $\endgroup$
    – Maksim
    Mar 20 at 14:33
  • $\begingroup$ I think it is not necessarily upper semicontinuous. Consider $\mu(B) = \infty$ if $0 \in B$ and $\mu(B) = m(B)$ otherwise where $m$ denotes the lebesgue measure. For any $r > 0$, $g(r) = \mu((0, 2r)) = 2r$, but $lim$ $sup_{y \to r}$ $g(y) = \infty$. $\endgroup$
    – Maksim
    Mar 20 at 15:13

$\{x:f(x)\geqslant a\}=\{x:{\operatorname{lim inf}}_{r\to 0}\frac{\mu((x-r,x+r))}{r}\geqslant a\}=\{x:\lim\limits_{\eta\to 0^+}\inf\limits_{0<r\leqslant\eta}\frac{\mu((x-r,x+r))}{r}\geqslant a\}$

$=\cap_{k=1}^{+\infty}\cup_{l=1}^{+\infty}\cap_{m=l}^{+\infty}\{x:\forall r\leqslant {1\over m},\frac{\mu((x-r,x+r))}{r}\geqslant a-\frac{1}{k}\}$.

Denote $A_{m,k}:=\{x:\forall r\leqslant {1\over m},\frac{\mu((x-r,x+r))}{r}\geqslant a-\frac{1}{k}\}$, we want to prove that it is a closed set.

Let $\{x_n\}$ be a sequence in $A_{m,k},x_n\to x^*$.WLOG,$x_n\leqslant x^*$.

$\forall \epsilon>0,\forall r\leqslant\frac1m$,let $\delta=x^*-x_N\leqslant \epsilon r$.

Since $(x_N-r+\delta,x_N+r-\delta)\subset (x^*-r,x^*+r)$,we have

$\frac{\mu((x^*-r,x^*+r))}{r}\geqslant \frac{\mu(x_N-r+\delta,x_N+r-\delta)}{r-\delta}{{r-\delta}\over r}\geqslant (a-\frac1k)(1-\epsilon)$.

So we have $x^*\in A_{m,k}$ because $\epsilon$ is arbitary.


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