Is this application of a law of large numbers rigorous in this not identically distributed case?

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Let $ \{ X_t \}_{t \in \mathbb{N} }$ be a sequence of indipendent random variables such that $X_t \sim N(u_t, 1)$ for all $t$ where the mean $u_t$ is given by the equation

$$u_t = \theta u_{t-1} + \epsilon_{t-1}$$

where $|\theta| < 1$, $\epsilon_t \sim N(0,1)$ for all $t$ and $\epsilon_t$ is independent from $\epsilon_s$ for all $t \ne s \in \mathbb{N}$. (we can assume a starting value for $u_0$).

Can one apply a law of large numbers of the following type:

Theorem (law of large numbers): Let $ \{ X_t \}_{t \in \mathbb{N} }$ be a sequence of indipendent random variables with standard deviation $\sigma_t$ and let $\lim_{t \rightarrow \infty} \sigma_t / t = 0$ then $$S_n := \frac{1}{n} \sum_{i=0}^n X_i \overset{p}{\to} \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=0}^n E[ X_i] $$

to obtain that

$$ S_n \overset{p}{\to} \lim_{n \rightarrow \infty} \frac{1}{n} \left[ u_0 \sum_{i=0}^n \theta^i + \epsilon_0 \sum_{i=0}^{n-1} \theta^i + \epsilon_1 \sum_{i=0}^{n-2} \theta^i + \dots + \epsilon_{n-1} \right] $$

and since by an analogous law of large numbers

$$ \epsilon_0 \sum_{i=0}^{n-1} \theta^i + \epsilon_1 \sum_{i=0}^{n-2} \theta^i + \dots + \epsilon_{n-1} \overset{p}{\to} 0 $$

we conclude that

$$ S_n \overset{p}{\to} \lim_{n \rightarrow \infty} \frac{1}{n} u_0 \sum_{i=0}^n \theta^i = \frac{u_0}{1-\theta} $$

Is this reasoning rigorous? if not, can it be made so?