Let $(f_{n})_{n}$ be a sequence of of functions in $L^{1}(\mu)$ ($\mu$ is $\sigma$-finite or finite ) such that it converges to $f$ in measure. Then, if $\lim_n\|f_{n}\|_{L^1}= \|f\|_{L^1}$, then $\lim_n \|f_n-f\|_{L^1}=0$.

Convergence in measure implies that there exists a subsequence $(f_{n_k})$ that $\mu$ almost surely converges to $f$, then $|f_{n_k}|$ converges to $|f|$, $\mu$ almost surely. In this case as $ \lim_k \|f_{n_k}\|_{L^1}= \|f\|_{L^1}$, $\lim_k \|f_{n_k}-f\|_{L^1}=0$. Now i stuck at showing that $\lim_n \|f_n-f\|_{L^1}=0$. Can we say that $\|f_n-f_{n_{k}}\|_{L^1}$ goes to $0$ as the index goes to infinity?

Edit : Not exactly sure if its true, but since $(f_{n_k})$ is also a sequence in $L^{1}(\mu)$, $f$ is in $L^{1}(\mu)$ due to completeness(?). I got two ideas after this point:

1)limit of $(f_{n})$ is finite hence $sup_{n} \int f_{n} <\infty$, so on finite measures $(f_{n})$ is uniformly integrable, which proves the result. Not sure about about the $\sigma$-finite measures.

2)Since $f_n$ converges, it is Cauchy(?) this proves the claim.

Edit 2: It seems that on $\sigma$-finite measure convergence in $L^{1}$ doesn't imply that $f$ is in $L^{1}$, unless the convergence of the sequence is almost uniformly. In finite measure the solution I proposed holds since almost everywhere implies almost uniform convergence by Egoroff's theorem. So the convergence of the subsequence implies that $f$ in $L^{1}$ and the rest is true.

  • $\begingroup$ You are not asked to show that $f$ \in $L^{1}(\mu)$. This is implicitly assumed. However if $||f_n||_1$ converges to some finite limit ( or it is just bounded) then $f$ \in $L^{1}(\mu)$. by Fatou's Lemma (applied to subsequences). $\endgroup$ – Kavi Rama Murthy May 10 '18 at 6:45

You know already everything, just choose the right sequence :)

Here comes the trick, consider the sequence $(||f_n-f||)$ and choose the subsequence $(n_k)$ that converges to $\limsup_n||f_n-f||$. All the assumptions on $(f_n)$ hold also for $(f_{n_m})$.

Now, you choose a subsubsequence $(n_{m_k})$ of $(n_m)$ as you have suggested in your question. An application of Scheffe's lemma reveals then that $||f_{n_{m_k}}-f||\rightarrow0$.

As $||f_{n_m}-f||\rightarrow\limsup_n||f_n-f||$, we find that $\limsup_n||f_n-f||=0$. Hence $||f_n-f||\rightarrow0$.

  • $\begingroup$ Would you mind explaining how we can pick such subsequence $(n_{k})$? the choice is not clear to me $\endgroup$ – dankmemer May 10 '18 at 1:24
  • $\begingroup$ Are you aware that $(||f_n−f||)=(a_n)$ is a sequence in the real numbers? That you can extract a sequence converging to the $\limsup$ is one of the properties of the $\limsup$ why mathematicians like the $\limsup$. Construction: assume $−\infty<\limsup a_n<\infty$ then for $n\in\mathbb{N}$ let $n_k$ be such that $|a_{n_k}−\sup_{m>n}a_m|<1/n$. This is possible, because of the definition $\limsup a_n=\lim_{n\rightarrow\infty}\sup_{m>n}a_m$. The case $\limsup a_n=\infty$ or $−\infty$ is a bit different but is also based on this definition of the $\limsup$. $\endgroup$ – Thomas Bernhardt May 10 '18 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.