If $\mu(X)<\infty$, $\int |f_n|<C$ for all n, and $f_n\rightarrow f$ a.e.Show that $f_n\rightarrow f$ in $L^1$.

Question

If $\mu(X)<\infty$, $\int |f_n|<C$ for all n, and $f_n\rightarrow f$ a.e.Show that $f_n\rightarrow f$ in $L^1$.

I tried to use uniform integrability but I could not figure out completely.

Does conditions $\int |f_n|<C$ and $\mu(X)<\infty$ imply also $f_n\rightarrow f$ in measure?

Thanks in advance,

Answer 1

Counterexample: Let $X=[0,1],f_n=2^n\cdot1_{[2^{-n},2^{-n+1})},f=0$.

Also, Egorov's Theorem gives you convergence in probability (almost sure convergence is stronger than convergence in probability in probability spaces).

In case $\int |f_n|^2d\mu < C$, observe that we can employ the Cauchy-Schwarz inequality to obtain $$ \int _E |f_n|d\mu \leq \mu(E)^{1/2} C^{1/2}. $$ for any set $E$. In particular $$ \int |f_n-f|d\mu\leq \int_{E^c} |f_n-f| d\mu + \mu(E)^{1/2}C^{1/2} + \int_E |f|d\mu. $$ Now use Egorov's Theorem to pick an appropriate set $E$.