Let's assume that $x$ is distributed according to some distribution, not necessarily Gaussian, with mean $\mu$ and standard deviation $\sigma$. Then, the distribution for the mean of a sample of $N$ iid ${x_i}$ converges to a Gaussian of mean $\mu$ and standard deviation $\sigma/\sqrt{N}$ if the following conditions hold:
- The ${x_i}$ are independent, although the convergence will still occur if the correlation is not too strong
- The convergence to a Gaussian still occurs if the ${x_i}$ have different pdf's, but their variances are of a similar order of magnitude
- The variances of the pdf's of the $x_i$ must be finite
- The theorem applies for $N\to \inf$. For finite $N$, the shape of the pdf of the sample mean will still be approximately Gaussian in its center (the tails will differ, though)
If each $x_i$ is a vector of dimension $n$, the situation is the same. The theorem can be applied to each dimension separately unless there are correlations between each vector's components.
In the continuous case, you say that $t$ is continuous, but $x$ is not necessarily continuous. Let's consider that $x(t)$ is piece-wise constant, that is, uniform over some intervals of $t$, but the value can differ from interval to interval. The width of each interval $dt_i$ can be in general a random variable too, where $i$ is the number of the interval. In this case, I expect that if the $\{dt_i\}$ fulfill the conditions mentioned above, and if there is no correlation between the $dt_i$ interval and the (uniform) value that $x$ takes in that interval, then the CLT should also apply.