# convergence in measure

• December 21st 2009, 12:34 AM
Kat-M
convergence in measure
Let $(X,F, \mu)$ be a measure space with $\mu (X) < \infty$.
Prove that $f_n \rightarrow f$ in measure $\Longleftrightarrow$ $\int_X \frac{|f_n-f|}{1+|f_n-f|} d \mu$ $\rightarrow 0$.
• December 21st 2009, 02:34 AM
Moo
Hello,

See here : http://www.mathhelpforum.com/math-he...-1-yn-0-a.html

(the integral is like the expectation, and measure = probability)
• December 21st 2009, 05:43 PM
Kat-M
i understand the first half. but i have a question on the second half.
by letting $Y_n=|f_n-f|$, $Z_n = \frac{Y_n}{1+Y_n}$and mimicking the steps. i have

$\int Z_n d\mu = \int Z_n 1_{\{Z_n \leq \epsilon \} } d\mu + \int Z_n 1_{ \{Z_n > \epsilon \} } d\mu$

$\int Z_n 1_{\{Z_n \leq \epsilon \} } d\mu \leq \int \epsilon 1_{\{Z_n \leq \epsilon \} } d\mu = \epsilon \int 1_{\{Z_n \leq \epsilon \} } d\mu$

but how do i show that $\epsilon \int 1_{\{Z_n \leq \epsilon \} } d\mu \leq \epsilon$ without knowing that $\mu$ is a probability measure? is there anyway i can estimate integral $\epsilon \int 1_{\{Z_n \epsilon \} }$? please help me.
• December 22nd 2009, 02:55 AM
Moo
It's not that difficult, think about an hypothesis you've not been using :D

Spoiler:
$\{Z_n<\epsilon\}\subset X \Rightarrow \mu(Z_n<\epsilon)=\int_X \bold{1}_{\{Z_n<\epsilon\}} ~d\mu \leq \mu(X) <\infty$

So you can say that it's $<\epsilon'$ where $\epsilon'=\epsilon/\mu(X)$