= Solution
<Convergence in probability> supplies a subsequence $X_{n_k}$ converging <almost surely> to $X$: choose it so $\Pr(|X_{n_k}-X|>2^{-k})<2^{-k}$, then apply the first <Borel-Cantelli lemma>. For each fixed $N$, the tail of this subsequence lies in $\sigma(X_n:n\ge N)$, and its almost sure limit is $X$. Thus $X$ has a version measurable with respect to the completed tail sigma-field of the independent sequence.
By <Kolmogorov's zero–one law>, every event in that tail sigma-field has <probability> zero or one. Consequently $\Pr(X\le r)\in\{0,1\}$ for every rational $r$. Since $X$ is a finite real <random variable>, its distribution function has limits zero and one at the two infinities; monotonicity then identifies a single threshold $c$ with $\Pr(X=c)=1$. Hence \b[$X$ is <almost surely> constant]. Completion only adds null events and does not alter the zero-one conclusion.
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