Variance of expectation estimator
Without more knowledge on $X$, we cannot say a lot. However, if it is IID with mean $\mu$ and variance $\sigma^2$, the sample mean $\hat{\mu} = \frac{1}{N}\left(\sum_{n=0}^{N-1}x_n\right)$ is unbiased, since:
$$E[\hat{\mu}] = E\left(\frac{1}{N}\sum_{n=0}^{N-1}x_n\right) = \frac{1}{N}\sum_{n=0}^{N-1}E\left(x_n\right) = N\mu/N=\mu\,.$$
And its variance goes to zero when $N$ increases:
$$V[\hat{\mu}] = V\left(\frac{1}{N}\sum_{n=0}^{N-1}x_n\right) = \frac{1}{N^2}\sum_{n=0}^{N-1}V\left(x_n\right) = N\sigma^2/N^2=\sigma^2/N\,.$$
Thus, the expectation converges to the actual mean, and the variance of the estimator tends to zero as the number of samples grows. Under these definitions, the sample mean is a consistent estimator. Note that one could try to use other hypotheses: alternative norms, convergence in law, etc.