One can also give a direct proof. Suppose $f:[0,1]\to \Bbb R$ is Riemann integrable.
Lemma Suppose that for a partition $P$; $\displaystyle D(f,P)=\sum_{k=0}^n (M_k-m_k)\Delta x_k<\varepsilon$. Then $M_k-m_k<\varepsilon $ for some $k$.
Proof Else $\displaystyle\varepsilon =\sum_{k=0}^n \varepsilon\Delta x_k\geqslant \sum_{k=0}^n (M_k-m_k)\Delta x_k$, proving the contrapositive.
Obs More generally, we get $M_k-m_k<(b-a)\varepsilon$ if the interval of integration is $[a,b]$.
We proceed to the proof of your claim.
Notation For a set $S$, we denote $o(f,S)=\sup\limits_{t,t'\in S}|f(t)-f(t')|$.
With this lemma at hand, we may start rolling: take $\varepsilon =1$. We know there exists a partition $P_1=\{0=x_0,x_1,\ldots,x_n=1\}$ with $D(f,P_1)<1$, since $f$ is Riemann integrable. By the lemma, there is a subinterval $I_i=[x_{i-1},x_i]$ where $o(f,I_i)<1$. If $i\neq 1$ or $n$, let $K_1=I_i$. Else, pick $x_0<x_0'<x_1$ or $x_{n-1}<x_n'<x_n$, and replace this in the endpoints of $K_i$. Now $f$ is integrable over $K_1$, so there exists a partition $P_2$ for which the sum in $K_1$, $D(f,P_2)<\frac 1 2$, which gives some subinterval $K_2$ where $o(f,K_2)<\dfrac 1 2|K_1|<\dfrac 1 2 $, since $|K_1|<1$. Again, we replace endpoints if necessary. Since we're shrinking the domain, in any case the oscillation descreases, so we're safe. Inductively, we can construct a sequence of strictly nested closed bouned intervals $$K_1\supsetneq K_2\supsetneq K_2\supsetneq\cdots$$ where $o(f,K_i)<\frac 1 i$. By Cantor, there is $x_0\in\bigcap\limits_{j\geqslant 1} K_j$. I claim $f$ is continuous at $x_0$. (Proof?)
Now consider $[0,x_0],[x_0,1]$. Note that by construction $x_0\neq 0,1$, so repeating this on $[0,x_0]$ we obtain $x_1\neq x_0$ where $f$ is continuous, and so on. Using this, we may also show
Claim The set where $f$ is continuous is dense in $[0,1]$.
Let $\chi_A$ be the characteristic function of the set $A$.
Then, for any $n\geq 1$,
- $\chi_{C_n}$ is a step function and its integral is equal to $\left(\dfrac{2}{3}\right)^n$ over $[0,1]$.
- $\chi_C\leq \chi_{C_n}$.
- the zero function is a step function lower than $\chi_C$ on $[0,1]$.
So, for any $n\geq 1$, $$0\leq \sup_{u\leq \chi_C} \int_0^1u \leq \inf_{u\geq \chi_C} \int_0^1u \leq \left(\dfrac{2}{3}\right)^n$$
where the $\sup$ and $\inf$ are taken over step functions on $[0,1]$.
Thus the upper integral $U(\chi_C)$ and the lower integral $L(\chi_C)$ of $\chi_C$ are equal and $\chi_C$ is integrable with $$\int_0^1\chi_C=0.$$
Best Answer
A Riemann integrable function has the property that its set of discontinuities has Lebesgue measure zero. By contrast, the fat Cantor set is nowhere dense and of positive measure, so its indicator function has the property that in a neighborhood of every point of the fat Cantor set, the function takes on values of 1 and 0. In particular the fat Cantor set is contained in the set of discontinuities of $\chi_C$.
Changing the values on any measure zero subset will not change the size of the set of discontinuities in this case. This follows from the fact that, up to removing a set of measure zero, $C$ will still be contained in the discontinuities set (it matters that $C$ is nowhere dense for this part). To see this, let $f = \chi_C$ $a.e.$ and let $M$ be the Lebesgue measure zero set on which we have changed the value of the function. Take $x \in C\cap M^c$ and observe that since $C$ is nowhere dense and since $M$ cannot contain an interval, there is a point of $C^c\cap M^c$ in any neighborhood of $x$ and so $x$ is a discontinuity point of $f$. But $\mu(C \cap M^c) = \mu(C)$ so $f$ cannot be Riemann integrable.