Given a locally integrable function $f: \mathbb R \to \mathbb R$, define $ K: \mathbb R \times \mathbb R+ \to \mathbb R$ by

$$ K(x, r) := \begin{cases} 1, & \text{if }f(x) > \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ -1, & \text{if }f(x) < \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ 0,& \text{if }f(x) = \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x \end{cases} $$

Define $U: \mathbb R \to [-1, 1]$ and $L: \mathbb R \to [-1, 1]$ by

$$ \begin{split} U(x) &:= \limsup_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t\\ L(x) &:= \liminf_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t \end{split} $$

Intuitively, $U$ and $L$ represent the weighted proportion of time a function spends above or below its average value in an infinitesimal neighbourhood of a point.

i) Is it true that for any locally integrable function, $U = L$ a.e?

ii) Is it true that $U = 1, 0, or -1$ a.e.?


1 Answer 1


For i) the Brownian motion gives a counter example.

We choose $f(t)=W_t$ with $W_t$ a Brownian on $\mathbb{R}$. Because one can calculte $U(0)$ and $L(0)$ from the Brownian motion restricted to any neighbourhood of $0$, we can apply Blumenthal’s 0-1 law: There exists $c_1,c_2\in \mathbb{R}$ such that $\mathbb{P}(U(0)=c_1)=1$ and $\mathbb{P}(L(0)=c_2)=1$. By symetry we have $c_1 =-c_2$. Moreover for any $0<a<1$ there exists $\epsilon$ such that for all $r>0$ $$\mathbb{P}\Big(\frac{1}{2r}\int_{B_r(0)}K(0,t)dt \geq a\Big)>\epsilon$$ Therefore $c_1\geq a$ and we conclude that $c_1=1$ and $c_2=-1$.

Finally by translation invariance of the Brownian motion we have for all $t\in \mathbb{R}$ $$\mathbb{P}(U(t)=1)=\mathbb{P}(L(t)=-1)=1$$

  • $\begingroup$ How do you show that for each $a\in(0,1)$ there is some $\epsilon[>0?]$ such that for all $r>0$ $P(\frac{1}{2r}\int_{B_r(0)}K(0,t)dt \ge a)>\epsilon$? Also, even if your have $P(U(t)=1)=P(L(t)=-1)=1$ for all $t$, how do you deduce from this that $P(U(t)=1\ \forall t)=1$? $\endgroup$ Apr 29, 2020 at 18:36
  • $\begingroup$ @losif Pinelis: The Brownian motion is scale invariant ($f^{(r)}(t):=\frac{1}{\sqrt{r}}W_{rt}$ is also a brownian motion). So $K(0,t)$ have the same law as $K(0,r^{-1}t)$. So $\mathbb{P}(\frac{1}{2r}\int K(0,t)dt\geq a)$ is independant of $r$. $\endgroup$
    – RaphaelB4
    Apr 30, 2020 at 7:16
  • $\begingroup$ @losif Pinelis : The second point you mention is not true for all t but almost all $t$. Calling $\Omega$ the probability space of the brownian motion then $\int_{-L}^L\int_{\Omega} 1_{U(t)=1}dt d\mu=2L=\int_{\Omega} \int_{-L}^L1_{U(t)=1}dt d\mu$. So with probability 1, for almost all $t\in [-L,L]$ $U(t)=1$ $\endgroup$
    – RaphaelB4
    Apr 30, 2020 at 7:20
  • $\begingroup$ All right, thank you. $\endgroup$ Apr 30, 2020 at 20:09

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