 Solution

Convergence in probability supplies a subsequence $X_{n_k}$ converging almost surely to $X$: choose it so $\mathrm{Pr}(|X_{n_k}-X|>2^{-k})<2^{-k}$, 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 $\mathrm{Pr}(X\le r)\in \{0,1\}$ 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 $\mathrm{Pr}(X=c)=1$. Hence $X$ is almost surely constant. Completion only adds null events and does not alter the zero-one conclusion.