# Liminf to limsup

• Nov 19th 2009, 08:40 AM
Liminf to limsup
Suppose that $\mid f_n \mid \leq g \in L^1$ and $f_n \rightarrow f$ in measure.

Then $\int f = \lim \int f_n$.

Proof so far.

Now, $g- f_n \geq 0$ and $f_n -g \geq 0$.

So for the first case, then

$\int g - f \leq lim \ inf \int g - f_n = \int g - \ lim \ sup \int f _n$

But how does the lim inf turn into lim sup?

Thank you!
• Nov 19th 2009, 09:55 AM
Jose27
What does "in measure" mean? If i'ts pointwise convergence a.e. with respect to the measure couldn't you apply Fatou two times (to $f_k$ and $-f_k$) to get that $\limsup \int f_k \leq f \leq \liminf \int f_k$

As for your question $\liminf (-f_k)=-\limsup (f_k)$
• Nov 20th 2009, 10:49 AM
Moo
Convergence in measure :

$\forall \epsilon >0, \lim_{n\to\infty} \mu(\{|f_n-f|>\epsilon\})=0$