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Suppose $a : \mathbb R_+ \to \{-1,1\}$ is a measurable function. Let $X_0 =\frac12$. Consider a particle that moves on the $X-$axis as follows. $$X_t = X_0 + \int_0^t a_s ds$$ where the integral is a Lebesgue integral.

Fix a $T=\frac12$. So, $X_t \in [0,1]$ for all $t \le T$.

Let $S \subset [0,1]$ be a set such that $\ell(S) =1$, where $\ell(\cdot)$ is the Lebesgue measure.

Define, $$G:= \{t \le T: X_t \in S\}.$$

Is it the case that $\ell(G) = \ell([0,T]) = \frac12$?

That is, the particle spends almost no time outside $S$?

avk255
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1 Answers1

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Yes, this is true, but the proof is not simple. First, you have to use the fact that the primitive of the Lebesgue integrable function $a$ is differentiable a.e. and its derivative is $a$. Hence, there is a set $E_0\subset[0,1/2]$ with $\ell(E_0)=0$ such that $X'(t)=a(t)\in\{-1,1\}$ for all $t\in [0,1/2]\setminus E_0$.

Then you have to use a result of Serrin and Varberg derivative zero that says that if you have a differentiable function, in this case $X:[0,1/2]\setminus E_0\to\mathbb{R}$ and you take a set of measure zero, $F=[0,1]\setminus S$, then $X'(t)=0$ for a.e. $t\in X^{-1}(F)$. Hence, there exists a set $E_1\subset[0,1/2]$ with $\ell(E_1)=0$ such that $X'(t)=0$ for all $t\in X^{-1}(F)\setminus E_1$. Since $X'(t)=a(t)\in\{-1,1\}$ for all $t\in [0,1/2]\setminus E_0$, it follows that $X^{-1}(F)\setminus E_1\subset E_0$. Thus, $X^{-1}(F)\subset E_0\cup E_1$, which has measure zero.

Gio67
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