I am reading "Analysis on Manifolds" by James R. Munkres.
Let $f:A\to Y$.
In this book, Munkres defines $\lim_{x\to x_0} f(x)$ only for $x_0$ which is a limit point of $A$.
So, if $x_0$ is an isolated point of $A$, we cannot compute $\lim_{x\to x_0} f(x)$.
There is the following theorem in this book:
Theorem 13.5(on p.109)
Let $S$ be a bounded set in $\mathbb{R}^n$; let $f:S\to\mathbb{R}$ be a bounded continuous function. Let $E$ be the set of points $\mathbf{x_0}$ of $\text{Bd}\,S$ for which the condition $$\lim_{\mathbf{x}\to\mathbf{x_0}} f(\mathbf{x})=0$$ fails to hold. If $E$ has measure zero, then $f$ is integrable over $S$.
And its proof is the following:
My question is here:
If $\mathbf{x_0}$ is an isolated point of $S$ and $f(\mathbf{x_0})\ne 0$, then $f_S$ is not continuous at $x_0$.
I think Munkres need to show that $\{\mathbf{x} \mid \mathbf{x} \text{ is an isolated point of }S\text{ and }f(\mathbf{x})\ne 0\}$ has measure zero in $\mathbb{R}^n$ to use the following theorem:
Let $Q$ be a rectangle in $\mathbb{R}^n$; let $f:Q\to\mathbb{R}$ be a bounded function. Let $D$ be the set of points of $Q$ at which $f$ fails to be countinuous. Then $\int_Q f$ exists if and only if $D$ has measure zero in $\mathbb{R}^n$.
Best Answer
There can only be a countable number of isolated points for any set in $\mathbb R^{n}$. And any countable set has measure $0$.