= Solution
For $k\geq1$, let $E_k$ be the set of numbers whose first digit $5$ occurs in position $k$. The first $k-1$ digits each have nine allowed values, while the $k$th is fixed, so $E_k$ is a union of decimal intervals of total length
$$
|E_k|=\frac1{10}\left(\frac9{10}\right)^{k-1}.
$$
The set $E_\infty$ of numbers having no digit $5$ can, after $N$ places, be covered by intervals of total length $(9/10)^N$; hence it has length zero.
For each fixed truncation level, the <function> is constant on finitely many decimal cylinders apart from the remaining set, whose covering length can be made arbitrarily small. The <Riemann integrability criterion> therefore shows that every truncation is Riemann integrable.
For integer $N$,
$$
\int_0^1\min(f,N)\,dx
=\sum_{k=1}^{N-1}
\frac{k}{10}\left(\frac9{10}\right)^{k-1}
+N\left(\frac9{10}\right)^{N-1}.
$$
The final term tends to zero, and monotonicity handles noninteger truncation levels. Therefore the <first occurrence of a decimal digit> calculation gives
$$
\int_0^1f(x)\,dx
=\sum_{k=1}^{\infty}
\frac{k}{10}\left(\frac9{10}\right)^{k-1}
=\frac{1/10}{(1-9/10)^2}
=\boxed{10}.
$$
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