What is $\bar\mu_n$? Am I right in thinking that it is the average of the sum of $n$ different populations means? 
Here it is used in the context that confused me. It's the Chebychev WLLN, apparently. 
"If $x_i, i = 1, . . ., n$ is a sample of $n$ observations such that $E[x_i] = \mu_i < \infty$ and Var[$x_i] = \sigma_i^2$ such that $\bar\sigma_i^2/n = (1/n^2)\Sigma_i \sigma_i^2 \rightarrow 0$ as $n \rightarrow \infty$ then $plim(\bar x_n - \bar\mu_n$) = 0."
Is this saying that each sample of $i$, corresponds to it's own population of $i$ (from above, $E[x_i] = \mu_i$) and as the sample get bigger we have to average over populations?
So if I were to draw the random variable 1 from a population of {1,2,3} and the random variable 4 from the population {4,5,6} then the population of my sample 1,4 is {1,2,3,4,5,6}?
 A: You are right.  
"If $x_i, i = 1, . . ., n$ is a sample of $n$ observations such that $E[x_i] = \mu_i < \infty$ and Var[$x_i] = \sigma_i^2$ such that $\bar\sigma_i^2/n = (1/n^2)\Sigma_i \sigma_i^2 \rightarrow 0$ as $n \rightarrow \infty$ then 
$$\lim_{n\rightarrow \infty}P\left(\left|\frac 1n\sum_{i=1}^nX_i - \frac 1n\sum_{i=1}^nE(X_i)\right|<\epsilon\right) =1$$
I guess you can make the notational mapping.
Since by design we assume different moments for each $X_i$, each comes from a different population. So if by $\{1,2,3\}$ you mean values of the index $i$, then $\{1,2,3\}$ is not a population, but a set including three values of the index with each value representing a different population.  
If you consider the random variables $\{X_1,X_4\}$, it is a pair coming from two different populations -you do not "unite" the two populations "into one" because, being different with respect to the object of study (convergence of sample moments), how could they form a single population (for the purposes of the specific study)? Have you contemplated how is the abstract concept of "statistical population" defined?
