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See discussion of "weak convergence" or "narrow convergence" of measures. Let $\mu_n$ be the measure with mass $1/n$ at each of $x_1,\dots,x_n$. Your condition B is what can be taken as the definition for: $\mu_n$ converges narrowly to $\mu$. The equivalent condition in terms of open sets is not your condition A, but rather $$\mu(O) \le \liminf_{N \to \infty} \frac{#(O,N)}{N} frac{\#(O,N)}{N}$$ for all open sets $O$.
See discussion of "weak convergence" or "narrow convergence" of measures. Let $\mu_n$ be the measure with mass $1/n$ at each of $x_1,\dots,x_n$. Your condition B is what can be taken as the definition for: $\mu_n$ converges narrowly to $\mu$. The equivalent condition in terms of open sets is not your condition A, but rather $$\mu(O) \le \liminf_{N \to \infty} \frac{#(O,N)}{N}$$ for all open sets $O$.