Question
Let $A_N = \left\{ \frac{1}{N}, \frac{2}{N}, \ldots, \frac{N}{N} \right\}$ and $A = \bigcup_N \left\{ \frac{1}{N} \right\} \times A_N$. Then we have
$$\overline{A} \simeq A \cup (\{0\} \times [0,1]).$$
We interpet $\overline{A}$ as a subset of $\mathbb{R}^2$ and endow it with the subspace topology.
Let $w_N \colon A_N \to \mathbb{R}$ with $\sum_{j = 1}^N w_N(j) = 1,\ w_N(j) \geq T$. This gives rise to a normalized, positive functional $w_N$ by
$$w_N(f) = \sum_{j \in A_N} w_N(j) f\left(N^{-1},j\right)$$
for any function $f \in C(\overline{A})$. So, $w_N$ corresponds to a probability measure. Assume that there is $x \in [0,1]$ such that
$$w_N(f) \to f(0, x)$$
for all bounded, continuous functions $f$ on $\overline{A}$. That is, the $w_N$ converge weakly to a Dirac measure at $x$.
Further, we are given a function $f \colon A \to \mathbb{R}$ such that $f(N^{-1},j_N) \to g(y)$ for all sequences $j_N \in A_N$ with $\frac{j_N}{N} \to y$. The function $y \mapsto g(y)$ is continuous. Therefore, $f$ has a unique continuous extension $\overline{f}$ to $\overline{A}$ (or am I missing here something?).
Thus, we can use the weak convergence of the functionals $w_N$ to obtain that
$$w_N(\overline{f}) \to \overline{f}(x).$$
Are there any mistakes?