# Relation between the measures of two sets defined via Lebesgue integration

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 $$S$$?

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.
• 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. – avk255 May 30 '19 at 9:06
• @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. – Mateusz Kwaśnicki May 30 '19 at 10:05
• 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? – avk255 May 30 '19 at 16:16