- #1
hwangii
- 5
- 0
Hi all,
I have a question about measure theory:
Suppose we have probability space [tex](\mathbb{R}^d,\mathcal{B}^d,\mu)[/tex] where [tex]\mathcal{B}^d[/tex] is Borel sigma algebra.
Suppose we have a function
[tex]u:\mathbb{R}^d\times \Theta\rightarrow \mathbb{R}[/tex] where [tex] \Theta\subset\mathbb{R}^l,l<\infty[/tex] and [tex]u[/tex] is continuous on [tex]\mathbb{R}^d\times \Theta[/tex].
Now consider the function [tex]G:\Theta\rightarrow[0,1][/tex] defined as follows:
[tex]G(\theta)=\int\limits_{\epsilon\in\mathbb{R}^d}\mathbb{I}\{u(\epsilon, \theta)\geq 0\} \mu (d\epsilon)[/tex]
where [tex]\mathbb{I}\{P\}[/tex] is an indicator function equal to 1 if P is true and 0 otherwise.
Is [tex]G(\theta)[/tex] continuous on [tex]\Theta[/tex]?
If you know the answer, could you please also tell me what kind of math books I need to look to find more about this? I would like to more about this by reading such text.
Thanks a lot!
I have a question about measure theory:
Suppose we have probability space [tex](\mathbb{R}^d,\mathcal{B}^d,\mu)[/tex] where [tex]\mathcal{B}^d[/tex] is Borel sigma algebra.
Suppose we have a function
[tex]u:\mathbb{R}^d\times \Theta\rightarrow \mathbb{R}[/tex] where [tex] \Theta\subset\mathbb{R}^l,l<\infty[/tex] and [tex]u[/tex] is continuous on [tex]\mathbb{R}^d\times \Theta[/tex].
Now consider the function [tex]G:\Theta\rightarrow[0,1][/tex] defined as follows:
[tex]G(\theta)=\int\limits_{\epsilon\in\mathbb{R}^d}\mathbb{I}\{u(\epsilon, \theta)\geq 0\} \mu (d\epsilon)[/tex]
where [tex]\mathbb{I}\{P\}[/tex] is an indicator function equal to 1 if P is true and 0 otherwise.
Is [tex]G(\theta)[/tex] continuous on [tex]\Theta[/tex]?
If you know the answer, could you please also tell me what kind of math books I need to look to find more about this? I would like to more about this by reading such text.
Thanks a lot!