Does the existence of the mean of a bounded, real-valued sequence imply it has a limiting distribution?

Choose $(x_n)$ and $(y_n)$ as bounded sequences for which

$$ \frac{1}{n}\sum_{k=1}^{n} \delta_{x_k} \xrightarrow[\text{as } n\to\infty]{d} \mu \qquad\text{and}\qquad \frac{1}{n}\sum_{k=1}^{n} \delta_{y_k} \xrightarrow[\text{as } n\to\infty]{d} \nu, $$

where $\mu$ and $\nu$ are probability measures on $\mathbb{R}$ having the same first moment. For example, we may choose $(x_n) = (0, 1, 0, 1, \ldots)$ and $(y_n) = (\frac{1}{2}, \frac{1}{2}, \ldots)$ so that $\mu = \frac{1}{2}(\delta_0 + \delta_1)$ and $\nu=\delta_{1/2}$.

Now let $(N_k)_{k\geq 1}$ be a sequence of positive integers such that $ N_{k+1}/N_k \to \infty$ as $k\to\infty$, and define

$$ (a_n)_{n\geq 1} = (x_1, \ldots, x_{N_1}) \oplus (y_1, \ldots, y_{N_2}) \oplus (x_{N_1+1}, \ldots, x_{N_3}) \oplus (y_{N_2+1}, \ldots, y_{N_4}) \oplus \cdots, $$

where $\oplus$ is the concatenation operation. Then it is not hard to show that $\frac{1}{n}\sum_{k=1}^{n} a_k$ converges to the common first moment of $\mu$ and $\nu$, but $\frac{1}{n}\sum_{k=1}^{n} \delta_{a_k}$ does not converge in distribution.