We know for $ \forall x \in [0,1], \exists \{a_n(x), n \in \mathbb{N} \} \subset \{0, 1\}$ with $ x = \sum_{n=1}^{\infty}\frac{a_n(x)}{2^n}$. (''binary expansion''). Show that $\{x \in [0,1]: \lim_{N\to \infty} \frac{1}{N} \sum_{n=1}^{N}a_n(x) = \frac{1}{2} \} \in {\mathcal{M}} \cap {\mathscr{R}}^c$. (i.e. Show it is measurable BUT not Riemann Lengthable.)
I think it is easy to show the measurability. We can let f(x) = $\lim_{N\to \infty} \frac{1}{N} \sum_{n=1}^{N}a_n(x)$, then it is enough to show $[f(x) = 1/2] \cap [0,1]$ is measurable. $[f(x) = 1/2] = f^{-1} (1/2)$ is measurable iff f is measurable. Let $f_N = \frac{1}{N} \sum_{n=1}^{N}a_n(x)$, then $f_N \rightarrow f$ ae, $f_N$ measurable thus f is measurable. Is it correct?
Hints: The sequence $a_n(x)$ is not necessarily unique because of the string-of-$1$'s phenomenon. For example $1/2 = .1000\bar 0 \cdots = .0111\bar 1 \cdots.$ Let's agree that for any such $x\in [0,1),$ we use the binary expansion ending in all $0$'s. (The set of such $x$ is countable, so it really doesn't matter which binary expansion we use in this case.)
That out of the way, there is a formula for each $a_n(x).$ Let $F$ be the floor function. Then
$$a_n(x) = F(2(2^{n-1}x - F(2^{n-1}x))), n \in \mathbb {N}, x\in [0,1).$$
Show that this proves the set in question is Borel measurable.
Let's call this set $E.$ To show $\chi_E$ is not Riemann integrable, argue that both $E,[0,1)\setminus E$ are dense in $[0,1).$