Common proofs on the law of large numbers usually assume a sequence of IID random variables. If $X_1\dots X_n$ has a common expected value $\mu$, finite but not necessarily common variance (hence not necessarily identically distributed), and are uncorrelated (so not necessarily independent), can the typical proof using Chebychev's Theorem be modified in the following way to show that $\lim_{n \to \infty} P(|\bar X_n-\mu|>\epsilon) = 0$:
If $\bar X_n = \frac{1}{n}\sum X_i$, since $X_1\dots X_n$ are uncorrelated, $Var(\bar X_n)=\frac{\sum\sigma_i^2}{n^2}$ and $E(\bar X_n)=\mu$. Chebychev's inequality gives:
$$P(|\bar X_n-\mu|>\epsilon) \leq \frac{Var(\bar X_n)}{\epsilon^2} = \frac{\sum\sigma_i^2}{n^2\epsilon^2}$$
As $n \to \infty$, the right hand side would approach zero as long as $n^2$ grows faster than $\sum\sigma_i^2$ (which would be the case if all $\sigma_i^2$ are equal). Does this show the WLLN?