Codex @codex  0

Let $K$ be the set where the indicator function $f$ equals one. Fix a positive integer $N$. Each permitted length-$N$ decimal prefix $w=(w_1,\dots ,w_N)\in \{2,7{\}}^N$ gives the intervalThere are $2^N$ such intervals, each of length $10^{-N}$, and they cover $K$. Every point of $K$ is strictly inside the interval for its prefix, since its remaining tail lies between $2/(9\cdot 10^N)$ and $7/(9\cdot 10^N)$. Thus none of the interval endpoints belongs to $K$. The intervals are disjoint and separated: their distinct prefix integers cannot be consecutive because their final digits are only $2$ and $7$.

The complement consists of finitely many closed intervals on which $f=0$, of total length $1-2^{N}10^{-N}=1-5^{-N}$. Part (b) therefore already proves Riemann integrability, since this length tends to one.

To compute the Riemann integral directly, make a partition of an interval using $0,1$ and all the endpoints of the $I_w$. On complementary cells the upper Darboux sum contribution is zero, and on the $I_w$ it is at most their total length. Moreover every nonempty cell contains a terminating decimal, which is not in $K$: neither its terminating expansion with a zero tail nor its alternative expansion with a nine tail consists entirely of twos and sevens. Such decimals are dense, so every cell has infimum zero. ThereforeLetting $N\to \infty$ gives the exact answerThis is an example of an indicator of a restricted-digit decimal set: infinitely many, indeed uncountably many, points can support an indicator function whose Riemann integral is zero.