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Let $A \subset \mathbb{R^n}$ be Jordan measurable and bounded. Let $I\subset \mathbb{R^n}$ be a closed and bounded interval such that $A \subset I$. We define the indicator function of $A$ as:

$\chi_A:I \rightarrow \mathbb{R}$, $x \, $$\mapsto$$ \, \chi_A(x)= \begin{cases} 1, & \text{$x \in A$} \\[2ex] 0, & \text{$x \notin A$} \end{cases}$

The set of points where $\chi_A$ is not continuous is clearly contained in $Bd(A)$. My question is, can there be points of $Bd(A)$ where $\chi_A$ is continuous?

Yagger
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Given any boundary point $x$ there exist one sequence $\{y_n\}$ contained in $A$ and converging to $x$ and another sequence $\{z_n\}$ contained in $A^{c}$ and converging to $x$. Along the first sequence the limit is 1 and along the second sequence the limit is 0, so $\lim_{u \to x} \chi_A (u)$ does not exist. Hence the function is not continuous at any boundary point.

Kavi Rama Murthy
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