Does a subsubsequence converging in measure implies that the sequence converges in measure?

Question

I want to prove

$D$ is a measurable set and $m(D)<\infty$. Two sequences of functions $\{f_n\}_{n\geq1}$ and$\{f_n\}_{n\geq1}$ are defined on $D$. $f_n$ converges to $f$ in measure and $g_n$ converges to $g$ in measure. Prove $f_ng_n$ converges to $fg$.

In the solution manual, the last step is that

Since the subsubsequecne $f_{n_{k_i}}g_{n_{k_i}} $ converges to $fg$ in measure, $f_ng_n$ converges to $fg$ in measure.

But I am not sure if this is correct.

By the way, the whole proof is

$\exists~\text{a subsequence}~ \{f_{n_k}\}$ of $\{f_n\}$ s.t. $f_{n_k}\rightarrow f$ a.e.,and $\exists~\text{a subsequence}~ \{g_{n_{k_i}}\}$ of $\{g_{n_k}\}$ s.t. $g_{n_{k_i}}\rightarrow g$ a.e. Then, $f_{n_{k_i}}g_{n_{k_i}} \rightarrow fg$ a.e.==> $f_{n_{k_i}}g_{n_{k_i}} \rightarrow fg$ in measure==>$f_ng_n \rightarrow fg$ in measure.

Answer 1

No, the convergence of a subsequence does not imply the convergence of the sequence in general.

However, if every subsequence has a subsubsequence that converges to the same limit $L$, then the whole sequence converges to $L$. (The proof given here works for any topological space.) This is what's being used here, as Daniel Fischer noted in a comment.

Indeed, it is impossible for a subsequence of $(f_ng_n)$ to converge a.e. to a function $h$ other than $fg$. The reason is that the corresponding subsequences of factors $f_n$, $g_n$ have subsubsequences that converge a.e. to $f$ and $g$, respectively.