We may take
$\alpha=\sum_n(1/73)^{f(n)}$
where $f(n)$ is any nonperiodic function.
The prime numbers $2$ and $3$ are both quadratic residues $\bmod 73$, hence products of powers of these primes will be so, also. So the numbers $(\alpha)(2^m3^n)$ will have nonuniformly distributed digits, leading to nonuniformly distributed values $\bmod 1$.
As an example, if we take $f(n)=n^2$ and sample $441$ numbers of the form $(\alpha)(2^m3^n)$ for $m,n$ from $0$ through $20$* (* -- MS Excel gives spurious zeroes due to roundoff error if we exceed $20$), we find $60$ between $0.0$ and $0.1\bmod 1$ and $54$ betwewn $0.9$ and $1.0$, with most of the other tenths between $35$ and $45$. We identify a peak in the distribution around $0.0$.
