1. ## convergence 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.

2. 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 L²), then the RHS goes to 0 too, and hence the convergence in measure.

3. Originally Posted by Moo
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 L²), then the RHS goes to 0 too, and hence the convergence in measure.
RHS goes to 0 because $\displaystyle \epsilon$ goes to 0, is it correct? how do i know that $\displaystyle \mu (|f-f_n >|\epsilon)$ goes to 0?

4. I'm rephrasing it

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

Since $\displaystyle f_n$ converges to $\displaystyle f$ in L², then by definition of the convergence in L², 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 ?

5. Originally Posted by Moo
I'm rephrasing it

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

Since $\displaystyle f_n$ converges to $\displaystyle f$ in L², then by definition of the convergence in L², 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 ?
yes thank you