= Solution
{c}
Let $\overline X_n=n^{-1}\sum_iX_i$ and write $\sigma^2=\operatorname{Var}(X_1)<\infty$. Independence gives $\mathbb E\overline X_n=\mu$ and $\operatorname{Var}(\overline X_n)=\sigma^2/n$. Apply part (b) to the nonnegative variable $(\overline X_n-\mu)^2$, which is the <Chebyshev inequality>:
$$
\mathbb P(|\overline X_n-\mu|>\varepsilon)
\leq\frac{\sigma^2}{n\varepsilon^2}\longrightarrow0.
$$
Thus \b[$\overline X_n\to\mu$ in <probability>], proving the finite-<variance> <weak law of large numbers>.
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