Let $(X, \mathcal{A}, \mu)$ be a measurable space, $(f_n)$ a sequence of integrable functions that converges in measure to some integrable function $f$. Assume further that there is some $g : X \rightarrow [0,\infty]$ integrable such that $|f_n|\leq g$ and $|f|\leq g$. Then $(f_n) \rightarrow f$ in mean.

I am not sure how to prove this. Since $(f_n)\rightarrow f$ in measure, every subsequence of $(f_n)$ admits some subsequence that converges to $f$ almost everywhere. I can prove that such a subsequence converges in measure, since convergence a.e plus being dominated by an integrable function as above implies convergence in mean.

If $(f_n)\rightarrow f$ a.e, then $(f_n-f)\rightarrow f$ a.e. Also, $$|f_n - f|\leq |f_n| + |f| \leq 2g$$ and therefore appealing to the Dominated Convergence Theorem we obtain that $\int_X |f_n -f|d\mu \rightarrow 0$ so $(f_n)\rightarrow f$ in mean.

The problem is then in showing that if a subsequence of a sequence $(f_n)$ converges in mean, then $(f_n) \rightarrow f$ in mean, and this is the part I need some help with. Thank you!

  • 1
    $\begingroup$ @MariosGretsas Absolutely! I've edited the statement. Thank you! $\endgroup$
    – user313212
    Commented Sep 5, 2017 at 10:44
  • $\begingroup$ If $(f_n)$ does not converge in mean to $f$, there is a subsequence, $(f_{n_k})_k$ and an $\epsilon>0$ so that $\Vert f_{n_k}-f\Vert_k >\epsilon$ for all $k$. $\endgroup$ Commented Sep 5, 2017 at 11:10

1 Answer 1


Assume that your statement is not true.

Thus exists an $\epsilon_0>0$ and a subsequence $f_{n_k}$ of $f_n$ such that $\int_X|f_{n_k}-f| \geq \epsilon_0,\forall k\geq 1$.

Since $f_{n_k}$ converges in measure then exists a subsequence $f_{n_{k_s}}$ of $f_{n_k}$ which converges to $f$ a.e

In other words $f_{n_{k_s}}-f \to 0$ a.e

We have that $|f_{n_{k_s}}| \leq g$ a.e thus $|f| \leq g$ a.e

Now from Lebesgue's Dominated Convergence theorem we have that $$\lim_{s \to +\infty}\int_X|f_{n_{k_s}}-f|=0$$ which is a contradiction.


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