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.

  • 3
    $\begingroup$ I guess the proof shows that every subsequence $(f_{n_k}g_{n_k})_{k}$ of $(f_ng_n)$ has a further subsequence $(f_{n_{k_i}}g_{n_{k_i}})$ that converges in measure to $fg$? $\endgroup$ Nov 4, 2015 at 13:29
  • $\begingroup$ If that is what the solution states, I would assume they are using a Cauchy sequence or this theorem I am forgetting about if we have a set $A$ that is closed and bounded then there exists a convergent subsequence, could be Bolzano-Weirstrass theorem but I can't remember $\endgroup$
    – Wolfy
    Nov 4, 2015 at 14:41
  • $\begingroup$ Well, actually the convergent subsequence $f_{n_k}$ is derived from F. Riesz Theorem (convergence in measure ==> some subsequence converges almost everywhere), so I guess that is not for every subsequence $f_{n_k}g_{n_k}$ but just for some subsequence $f_{n_k}g_{n_k}$. $\endgroup$
    – Xiaobo Yu
    Nov 4, 2015 at 16:07
  • $\begingroup$ This idea of passing to a subsequence or a subsubsequence is frequently used, but it depends on context to see how to make it work. There is never a general principle "$\gamma_{n_k}$ converges implies $\gamma_n$ converges" in any sense. Take a sequence $\gamma_n$ that does not converge (in whatever sense you like) and then augment it by adding in another sequence $\gamma'_n$ that does converge. Then $\gamma_1$, $\gamma'_1$, $\gamma_2$, $\gamma'_2$, ... will produce a bad sequence with a good subsequence. $\endgroup$ Nov 4, 2015 at 17:43

1 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.


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