= Solution
Let $K$ be the set where the <indicator function> $f$ equals one. Fix a positive integer $N$. Each permitted length-$N$ decimal prefix $w=(w_1,\ldots,w_N)\in\{2,7\}^N$ gives the interval
$$
I_w=[t_w,t_w+10^{-N}],\qquad t_w=\sum_{j=1}^Nw_j10^{-j}.
$$
There 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 $2/(9\cdot10^N)$ and $7/(9\cdot10^N)$. 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 $1-2^N10^{-N}=1-5^{-N}$. 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. Therefore
$$
0=L(f,P_N)\le\int_0^1f(x)\,dx\le U(f,P_N)\le5^{-N}.
$$
Letting $N\to\infty$ gives \b[the exact answer]
$$
\boxed{\int_0^1f(x)\,dx=0.}
$$
This 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.
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