Let $K \subset [0,1]$ be the Cantor set. show that $f : [0, 1] \to \mathbb R$ given by $$f(x) = \begin{cases} 1, & x \in K \\ 0, & x \in [0, 1]\setminus C \end{cases}$$ is integrable and calculate $\int_0^1 f(x)dx$.
My attempt:
The Cantor set $K$, although not countable, has a null measure. Thus, if we stop $n$-th step of its construction, we will see that the Cantor set is contained in the collection of $2^n$ intervals, each having length $\frac{1}{3^n}$.Given $ \epsilon \gt 0$, we can take $n \in \mathbb N$ such that $\frac{2}{3^n} \lt \epsilon$ and we conclude that the measure of $K$ is zero.
Now, as [ 0.1] is closed then $[0,1] - K$ is open, the function $f$ is locally constant, and therefore continues, at points $x \notin K$. Since $K$ has no interior points, $f$ is discontinued at all $K$ points. By the theorem(If every set D of the discontinuity points of a bounded function $f:[a,b] \to \mathbb R$ has null measure then $f$ is integrable) $f$ is integrable.
Given any partition $P$ of [0,1] all ranges of $P$ contain points that do not belong to $K$, since int $K = \emptyset$. Thus, $m_i=0$ and $s(f;p)=0$ for every partition $P$. It then follows that $\int_0^1 f(x)dx = \underline\int_0^1f(x)dx = 0$.
Thank's in advance for any help.