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

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.
 A: 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 $A \neq B$.

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 $A = 1$, $B = 0$.

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 $B \leqslant A$.
A: 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...
