Can Limits and Simple Functions Approximate Extended Real Functions?

[wayneckm]

Hello all,


I have a few questions in my mind:

1) $$\lim_{n\rightarrow \infty}[0,n) = \cup_{n\in\mathbb{N}}[0,n) = [0,infty)$$ holds, and for $$\lim_{n\rightarrow \infty}[0,n] = \cup_{n\in\mathbb{N}}[0,n] = [0,infty)$$ is also true? It should not be $$[0,infty]$$, am I correct?

2) Consider an extended real function $$f$$, if we use simple function $$f_{n} = f 1_{f\leq n}$$, by taking limit, we can only have it approximated to $$f 1_{f < \infty}$$ but since $$f$$ may take $$\infty$$, such simple function may not be approximating $$f$$ almost everywhere unless $$f = \infty$$ is of measure 0?

Am I correct? Thanks.


Wayne

 

[mathman]

Your first statement is correct. I can't figure out the second question - the symbol 1 after f in the expression f_n= means what?

 

[Landau]

Help me out here. What does
$$\lim_{n\rightarrow \infty}[0,n)$$
mean? Or were you just defining it as
$$\lim_{n\rightarrow \infty}[0,n) := \cup_{n\in\mathbb{N}}[0,n)\?$$

 

[wayneckm]

the symbol 1 here means the indicator function.

Thanks.