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The answer to the first question is yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$