Lemma about the integral of a function with compact support Lemma 16.4 (p. 140) of Munkres' Analysis on Manifolds says:

Let $A$ be open in $\mathbb{R}^n$; let $f: A \rightarrow \mathbb{R}$ be continuous. If $f$ vanishes outside a compact subset $C$ of $A$, then the integrals $\int_A f$ and $\int_C f$ exist and are equal.

The first step in his proof is saying that the integral $\int_C f$ exists because $C$ is bounded and $f$ is continuous and bounded on all of $\mathbb{R}^n$.
But don't you need $C$ to be rectifiable (i.e. bounded and boundary has measure $0$) for integrability? The fat cantor set is compact but not rectifiable, so the integral over it won't exist.
 A: I think that Theorem 13.5 of the same book may be of use here. The conditions set out in the lemma imply that $f$ is continuous at $\partial C$. Since it vanishes outside $C$, the limit must be $0$ at all points on $\partial C$. Therefore, $f$ and $C$ satisfy the conditions of Theorem 13.5.
It is a bit strange that Munkres didn't offer more explanation, seeing that he is far from terse for most of the prior material. On the other hand, the argument above could probably be considered standard at this stage of the book.
A: That depends on the theory of integration that you have at your disposal. Integration based on Jordan measure (i.e., multiple Riemann integral) indeed runs into trouble when sets have "fat" boundaries. But for Lebesgue integral the boundary is not an issue. Since every Borel set is Lebesgue measurable, we can integrate any bounded Borel function over any bounded Borel set. This covers the case of $f$ being integrated over $C$. 
A: I am reading "Analysis on Manifolds" by James R. Munkres.
My answer is the same as Shavak Sinanan's answer.
But I will write the statement of Theorem 13.5.

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 $x_0$ of the boundary of $S$ for which the condition $\lim_{x \to x_0} f(x) = 0$ fails to hold.
If $E$ has measure zero, then $f$ is integrable over $S$.

Since $C$ is compact, $C$ is a bounded set in $\mathbb{R}^n$.
Since $f$ is continuous on $A$ and $C\subset A$, $f\vert_C$ is continuous on $C$.
Since $C$ is compact and $f\vert_C$ is continuous on $C$, $f\vert_C$ attains a maximum value and a minimum value on $C$.
So, $f\vert_C:C\to\mathbb{R}$ is a bounded continuous function.
Let $x_0\in \operatorname{Bd}C$.
If $x_0$ is not an isolated point of $C$, then $\lim_{x\to x_0}f\vert_C(x)=\lim_{x\to x_0}f(x)=f(x_0)$.
Since $f$ vanishes outside $C$ and any neighborhood of $x_0$ contains a point in $A-C$, $f(x_0)=0$.
So, $\lim_{x\to x_0}f\vert_C(x)=0$.
So, the set of points $x_0$ of the boundary of $C$ for which the condition $\lim_{x \to x_0} f\vert_C(x) = 0$ fails to hold is $\emptyset$.
Since $\emptyset$ has measure zero, $f\vert_C$ is integrable over $C$ by Theorem 13.5.
So, $\int_C f\vert_C$ exists.
