# Convergence in measure and norm convergence implies convergence in norm

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.

• 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). – Kavi Rama Murthy May 10 '18 at 6:45

## 1 Answer

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 $||f_n-f||\rightarrow0$.

• Would you mind explaining how we can pick such subsequence $(n_{k})$? the choice is not clear to me – dankmemer May 10 '18 at 1:24
• 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 $\limsup$. – Thomas Bernhardt May 10 '18 at 9:54