convergence in measure

**Kat-M** · December 20th 2009, 11:34 PM

Let $(X,F, \mu)$ be a measure space with $\mu (X) < \infty$ .
Prove that $f_n \rightarrow f$ in measure $\Longleftrightarrow$ $\int_X \frac{|f_n-f|}{1+|f_n-f|} d \mu$ $\rightarrow 0$ .

**Moo** · December 21st 2009, 01:34 AM

Hello,

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

(the integral is like the expectation, and measure = probability)

**Kat-M** · December 21st 2009, 04:43 PM

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.