# a function defined by inferior limit related to Borel measure is Borel measurable

Question:

$$\mu$$ 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!

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 $$\{x:g(x) is open for any $$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.

• Fix an r, we suppose that $μ((x−r,x+r)) < \infty$, but why is it also true that $μ((x−2r,x+2r))< \infty$? Mar 20 at 4:20
• @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). Mar 20 at 14:12
• 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. Mar 20 at 14:33
• 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 $lim$ $sup_{y \to r}$ $g(y) = \infty$. 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

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

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.