I posted this question on StackExchange, people have upvoted it but I have not received any response. I read up here that it is okay to post unanswered StackExchange questions on Mathoverflow. So, posting it below.

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 $S$?

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Yes, by the coarea formula. In fact it is sufficient to assume that $a(t)$ is bounded and non-zero almost everywhere.

The function $X(t)$ is Lipschitz continuous (with Lipschitz constant 1), $g(t) = \mathbb{1}_{[0,T] \setminus G}(t)$ is integrable (in fact, bounded by $0$ and $1$), the zero-dimensional Hausdorff measure is just the counting measure, and thus $$ \begin{aligned} \int_0^T \mathbb{1}_{[0,T] \setminus G}(t) |X'(t)| dt & = \int_0^T g(t) |X'(t)| dt \\ & = \int_0^1 \biggl(\sum_{t \in [0, T] : X(t) = x} g(t)\biggr) dx \\ & = \int_0^1 \# \{t \in [0, T] : X(t) = x, t \notin G\} dx \\ & = \int_0^1 \# \{t \in [0, T] : X(t) = x, X(t) \notin S\} dx \\ & = \int_0^1 \# \{t \in [0, T] : X(t) = x, x \notin S\} dx \\ & = \int_{[0, 1] \setminus S} \# \{t \in [0, T] : X(t) = x\} dx = 0 .\end{aligned} $$ The interand $\mathbb{1}_{[0,T] \setminus G}(t) |X'(t)|$ is non-negative, and hence it is equal to zero almost everywhere. However, $X'(t) = a(t) \ne 0$ almost everywhere, and hence $[0,T] \setminus G$ is necessarily a null set.

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  • $\begingroup$ Thanks a lot! I need to understand this answer which will take some time I guess because I do not know the Coarea formula you cited. One thing I don't quite understand is the step $ \int_{[0, 1] \setminus S} \# \{t \in [0, T] : X(t) = x\} dx = 0$. Why is the integrand $0$? Also, I do not quite understand why $\int_0^T g(t) X'(t) dt = \int_0^1 \biggl(\sum_{t \in [0, T] : X(t) = x} g(t)\biggr) dx $. I am sure I am missing something simple. $\endgroup$ – avk255 May 30 '19 at 9:06
  • $\begingroup$ @avk255: Sorry, I was typing on a rush, there were some errors in my answer. Now everything should be OK. Regarding your questions: (1) $[0, 1] \setminus S$ has zero Lebesgue measure (by assumption), so the integral is zero. (2) This is precisely the coarea formula (although there was an absolute value missing): the sum is simply the integral over the zero-dimensional Lebesgue measure. $\endgroup$ – Mateusz Kwaśnicki May 30 '19 at 10:05
  • $\begingroup$ Thanks. I am going over your answer but it will take time for me to process it! What I am worried about is the $\# \{t \in [0, T] : X(t) = x\} $ may be $\infty$ no? $\endgroup$ – avk255 May 30 '19 at 16:16
  • $\begingroup$ Yes, of course. Still, the domain of integration is a null set, and thus the integral vanishes. Feel free to ask if you have any questions, I often make dull mistakes. $\endgroup$ – Mateusz Kwaśnicki May 30 '19 at 17:14
  • $\begingroup$ I think I have understood your answer. Thanks a lot! $\endgroup$ – avk255 Jun 2 '19 at 8:11

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