Let $X_1, X_2, \dots , X_n$ be normally distributed independent observations with known variance $\sigma^2$ and mean respectively given by $\mu_i = \mu + \epsilon_i$ where $\epsilon_i$ is white noise, i. e., $E[\epsilon_i] = 0$, $E[\epsilon_i^2] = \sigma_{\epsilon}$ and the $\epsilon_i$ are all symmetrically identically distributed and independent from one another.
We can see that the maximum likelihood estimator for $\mu$ remains consistent by maximizing the log-likelihood function
$$ \sum_{i=1}^n \log \frac{1}{\sqrt{2 \pi \sigma^2}} - \frac{(X_i - \mu - \epsilon_i)^2}{2 \sigma^2}$$
taking the derivative wrt $\mu$ and setting the result equal to zero one obtains that
$$ +2 \hat{\mu} = \frac{2}{n} \left( \sum_{i=1}^n X_i + \sum_{i=1}^n \epsilon_i \right)$$
by the strong law of large numbers $\sum_{i=1}^n X_i \rightarrow \mu$ and $\sum_{i=1}^n \epsilon_i \rightarrow 0$ almost surely. So from the linearity of the limit we obtain the consistency.
Is this true even for the variance? that is; if we had $X_1, X_2, \dots , X_n$ be normally distributed independent observations with known mean $\mu$ and variance respectively given by $\sigma_i^2 = \sigma^2 + \epsilon_i$ would the maximum likelihood estimator for $\sigma^2$ still be consistent ?