Let see $A \subset \mathbb{R}^n$, Define $\overline{f}(x) = \limsup_{y\to x} f(y)$ and $\underline{f}(x) = \liminf_{y\to x} f(y)$
$\underline{\chi_A} =\chi_{A^o}, \text{ }\overline{\chi_A} =\chi_{A^-} $
I wanna prove that. How can I approach?
Let see $A \subset \mathbb{R}^n$, Define $\overline{f}(x) = \limsup_{y\to x} f(y)$ and $\underline{f}(x) = \liminf_{y\to x} f(y)$
$\underline{\chi_A} =\chi_{A^o}, \text{ }\overline{\chi_A} =\chi_{A^-} $
I wanna prove that. How can I approach?
If $x \in A^{\circ}$ there exist $B_\varepsilon(x) \subset A$. Then $\inf \{\chi(x): x \in B_\varepsilon(x) - \{x\} \}= 1 = \chi(x)$. If $x \notin A^{\circ}$ there exist $z \in B_\varepsilon(x)- \{x\}$ such that $x \notin A^{\circ}$ and $\inf\{\chi(x): x \in B_\varepsilon(x)- \{x\} = 0 = \chi(x)$. Hence $ \underline{\chi_{A}} = \chi_{A^{\circ}}$.