Theorem (Law of large numbers): Let $\x, \x_1, \x_2, \ldots$ be independently and identically distributed random vectors such that $\Ex\norm{\x}^2 < \infty$. Then $\bar{\x}_n \inQM \bmu$, where $\bmu = \Ex(\x)$.
Proof:
\[\begin{alignat*}{2} \Ex\norm{\bar{\x} - \bmu}_2^2 &= \Ex\{ (\bar{\x} - \bmu) \tp (\bar{\x} - \bmu)\} \\ &= n^{-2} \Ex\left\{ \sum_{i,j} (\x_i - \bmu) \tp (\x_j - \bmu)\right\} &\hspace{8em}& \bar{\x} - \bmu = \tfrac{1}{n}\sum_i (\x_i-\bmu) \\ &= n^{-2} \sum_{i,j} \Ex\left\{ (\x_i - \bmu) \tp (\x_j - \bmu)\right\} && \textnormal{Linearity of expectation} \\ &= n^{-2} \sum_{i} \Ex\left\{ (\x_i - \bmu) \tp (\x_i - \bmu)\right\} && \Ex\left\{ (\x_i - \bmu) \tp (\x_j - \bmu)\right\} = \trace\{\textnormal{Cov}(\x_i, \x_j)\}, \\ & && \textnormal{which is 0 if } \x_i \ind \x_j \\ &= n^{-1} \Ex\left\{ (\x - \bmu) \tp (\x - \bmu)\right\} && \Ex \x_i, \Var \x_i \textnormal{ same for all } i \\ &= n^{-1} \Ex\left\{ \x \tp \x - 2 \bmu \tp \x + \bmu \tp \bmu \right\} \\ &\to 0 && \Ex\norm{\x}^2 \le \infty \end{alignat*}\]Note that this proof, unlike the usual law of large numbers, does not actually require $\lbrace \x_i \rbrace$ to be independent or identically distributed, only that the terms in the sum are uncorrelated and have the same mean and variance.