Prove an equivalent definition of $\sup\{T>0:x\in\overline{\mathbb R^n\setminus D_t },t \in [0,T) \},D_t\subset\mathbb R^n,x\in\partial D_0$

Question

Let $D_t$ be closed, bounded and non-empty subsets of $\mathbb{R}^n$ for $t \ge 0$. Let $x \in \partial D_0$ and define $$\mathcal{A}= \sup \{T>0: x \in \overline{\mathbb{R}^N \setminus D_t } \text{ for almost every } t \in [0,T) \}.$$ Assume that such sup exists and is finite. Then let $$\mathcal{B}= \lim_{\delta \to 0^+} \inf\{t>0: D_t \cap B_\delta(x) \not\subset D_0 \cap B_\delta(x) \},$$ where $B_\delta(x)$ is the ball of radius $\delta$ and centre $x$.

How can I prove that $\mathcal{A} = \mathcal{B}$?

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Assume also that if $D_0 \subset B_{R_0}(0)$, then $$D_t \subset B_{R_0 + C t^{\alpha}}\left(0\right),$$ where $C>0$ is fixed (possibly large) and $\alpha<1$ is fixed.

Answer 1

To make the definition of $A$ more clear, observe that $x\notin\overline{\mathbb{R}^{n}\setminus D_{t}}$ iff $x\in D_{t}^{o}$. For, $(\overline{\mathbb{R}^{n}\setminus D_{t}})^{c}=D_{t}^{c-c}=D_{t}^{o}$, where $D_{t}^{o}$ deontes the interior of $D_{t}$. Note that it is false that $x\in D_{0}^{o}$, so...

Answer 2

The conclusion $A = B$ is still incorrect.

Counter-example: Take$$ D_t = \begin{cases} \overline{B_2(0)}; & t = 1 \text{ or } t \geqslant 2\\ \overline{B_1(0)}; & \text{otherwise} \end{cases}, \quad x = (1, 0, \cdots, 0), $$ then for any $t \in [0, 1) \cup (1, 2)$, there is $x \not\in D_t^\circ$, i.e. $x \in \overline{\mathbb{R}^n \setminus D_t}$. Thus$$ A = \sup\{T > 0 \mid x \in \overline{\mathbb{R}^n \setminus D_t} \text{ for almost every } t \in [0, T)\} = 2. $$

However, for any $0 < δ < 1$, there is $$D_1 \cap B_δ(x) = B_δ(x) \not\subseteq D_0 \cap B_δ(x),$$ and$$ D_t = D_0 \Longrightarrow D_t \cap B_δ(x) = D_0 \cap B_δ(x), \quad \forall 0 \leqslant t < 1$$ thus$$ B = \lim_{δ \to 0} \inf\{t > 0 \mid D_t \cap B_δ(x) \not\subseteq D_0 \cap B_δ(x)\} = 1. $$ So $A \neq B$.

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Even if the definition of $A$ is changed as$$ A = \sup\{T > 0 \mid x \in \overline{\mathbb{R}^n \setminus D_t}, \forall t \in [0, T)\}, $$ it is still incorrect.

Counter-example: Take$$ D_t = \begin{cases} \overline{B_1(0)}; & t = 0\\ [-1, 1]^n; & 0 < t \leqslant 1\\ \overline{B_2(0)}; & t > 1 \end{cases}, \quad x = (1, 0, \cdots, 0), $$ then $A = 1$, $B = 0$.

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Now, to prove that $A \geqslant B$ in general, it suffices to use the new definition of $A$.

For any $ε > 0$, by definition there exists $A \leqslant t_0 < A + ε$ such that $x \not\in \overline{\mathbb{R}^n \setminus D_{t_0}}$, i.e. $x \in D_{t_0}^\circ$, which implies that there exists $δ_0 > 0$ such that $B_{δ_0}(x) \subseteq D_{t_0}^\circ$. Thus for any $0 < δ \leqslant δ_0$,$$ D_{t_0} \cap B_δ(x) = B_δ(x) \not\subseteq D_0 \Longrightarrow D_{t_0} \cap B_δ(x) \not\subseteq D_0 \cap B_δ(x), $$ which implies$$ \inf\{t > 0 \mid D_t \cap B_δ(x) \not\subseteq D_0 \cap B_δ(x)\} \leqslant t_0 < A + ε, $$ and$$ B = \lim_{δ \to 0^+} \inf\{t > 0 \mid D_t \cap B_δ(x) \not\subseteq D_0 \cap B_δ(x)\} \leqslant A + ε. $$ Make $ε \to 0^+$, then $B \leqslant A$.