$\displaystyle f_n \rightarrow f$ in $\displaystyle L^2$. Show that $\displaystyle f_n \rightarrow f$ in measure.

any hint would be appreciated.

Printable View

- Dec 31st 2009, 12:24 PMArchiconvergence in measure
$\displaystyle f_n \rightarrow f$ in $\displaystyle L^2$. Show that $\displaystyle f_n \rightarrow f$ in measure.

any hint would be appreciated. - Dec 31st 2009, 12:36 PMMoo
Hello,

$\displaystyle \forall \epsilon>0,\int (f_n-f)^2 ~d\mu \geq \int_{\{|f_n-f|>\epsilon\}} (f_n-f)^2 ~d\mu>\epsilon^2 \mu(|f_n-f|>\epsilon)\geq 0$

and since the LHS goes to 0 (convergence in Lē), then the RHS goes to 0 too, and hence the convergence in measure. - Dec 31st 2009, 02:10 PMArchi
- Jan 2nd 2010, 12:11 AMMoo
I'm rephrasing it (Doh)

$\displaystyle \forall \epsilon>0,{\color{red}\int (f_n-f)^2 ~d\mu} \geq \int_{\{|f_n-f|>\epsilon\}} (f_n-f)^2 ~d\mu>{\color{blue}\epsilon^2 \mu(|f_n-f|>\epsilon)}\geq 0

$

Since $\displaystyle f_n$ converges to $\displaystyle f$ in Lē, then by definition of the convergence in Lē, the red part goes to 0 as**n goes to infinity**

Thus by the sandwich theorem, the blue part goes to 0 as n goes to infinity.

When it is said "for all epsilon", it's like you choose any epsilon and you**fix**it. So it's like a constant !

Thus $\displaystyle \lim_{n\to\infty} {\color{blue} \mu(|f_n-f|>\epsilon)}=0$, which means, by definition, that $\displaystyle f_n$ converges to $\displaystyle f$ in measure.

Looks better ? - Jan 2nd 2010, 12:44 AMArchi