Show that there is no riemann integrable function $f$ on $[0,1]$ such that $f=\chi_{C}$ a.e. (almost everywhere), where $C$ is the fat cantor set.
Proof:
Would it suffice to show that $\chi_C$ is not riemann integrable??? Or am I missing something? Any suggestions?
Attempt:
Let $x$ be an element of $C$ then since $C$ contains no intervals if we define $C_n = (x-\frac{1}{n}, x+\frac{1}{n})$ then there exists some $x_n$ in $C_n$ such that $x_n\neq x$ and $x_n\not\in C$. Then $\forall \delta>0,\exists n$ such that $\delta>\frac{1}{n}.$ So then $\left|x_n-x\right|<\frac{1}{n}<\delta.$ But, if we take $\epsilon=1/2$ we get $|f(x_n)-f(x)|=1>\epsilon$ so $f$ is discontinuous at every $x\in C$, thus it cannot be Riemann integrable.
Would this work??