= Solution
{c}
Suppose first that <complete convergence> holds. For each positive integer $j$, apply the <First Borel-Cantelli lemma> to $\{|X_n-X|>1/j\}$. Outside a countable union of null events, only finitely many of these exceedances occur for every $j$. This is <almost sure convergence>, with no independence assumption.
For the converse, an important point is that independence of $X_n$ does not immediately imply independence of $X_n-X$. Instead use the fact that an <almost-sure limit of independent random variables is constant>. Indeed, $Z=\limsup_nX_n$ is measurable for the <tail sigma-algebra>. The <Kolmogorov zero-one law> makes each event $\{Z\leq q\}$, $q\in\mathbb Q$, have probability zero or one. Since $Z=X$ almost surely and $X$ is finite, these probabilities form the distribution function of a point mass: there is a finite constant $c$ with $X=c$ almost surely.
Now $\{|X_n-c|>\epsilon\}$ are independent events. <Almost sure convergence> says that their infinitely-often event has probability zero. The contrapositive of the <Second Borel-Cantelli lemma> gives
$$
\sum_n\mathbb P(|X_n-c|>\epsilon)<\infty.
$$
Replacing $c$ by the almost surely equal $X$ leaves each probability unchanged. Thus \b[complete convergence and almost sure convergence are equivalent for independent sequences].
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