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It is easy to construct $A$ for which the limit does not exist. Consider the following set $A$. Include all even numbers in the intervals $[10^n, 10^{n+1}]$ for even $n=0,2,...$, and all numbers divisible by 3 in the intervals $(10^n,10^{n+1})$ for odd $n$. Now if $N=10^k$ with $k$ even then the probability of 0 is $\ge .9$, and if $k$ is odd, then the probability of $1$ is $\ge .9*1/3=.3$. In general, the limit exists if the intervals between consecutive numbers in $A$ are "uniformly spaced".

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