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IN Rogers & Williams "Diffusions, Markov Process and Martingales" they introduce the resolvent as:

$$R_\lambda f(x):=\int_{[0,\infty)}e^{-\lambda t}P_tf(x)dt=\int_ER_\lambda(x,dy)f(y)$$

where $P_t$ is a measurable transition function, $R_\lambda(x,\Gamma):=\int_{[0,\infty)}e^{-\lambda t}P_t(x,\Gamma)dt,\Gamma\in \mathcal{E}$ and $f\in b\mathcal{E}$ of a measurable space $(E,\mathcal{E})$. In addition we can suppose that the map $(t,x)\mapsto P_t(x,\Gamma)$ is $\mathcal{E}\times \mathcal{B}([0,\infty))$ measurable. Right after these equations they write: "Trivial application of monotone-class theorems and of Fubini's theorem are now made without comment."

For the equation above you have to use Fubini. Furthermore, I think for the claim $R_\lambda:b\mathcal{E}\to b\mathcal{E}$ you have to use a monotone class argument. Using measuretheoretic induction, we can prove that $P_t:b\mathcal{E}\to b\mathcal{E}$. So $P_tf$ is again bounded and measurable. To conclude we must show that

$$\int_{[0,\infty)}e^{-\lambda t}P_tf(\cdot)dt\in b\mathcal{E}$$

given $P_tf(\cdot)\in b\mathcal{E}$. However here my problem starts. I need to find a class $\mathcal{H}$ of bounded functions on a set $S$ into $\mathbb{R}$ such that:

  1. $\mathcal{H}$ is a vector space over $\mathbb{R}$
  2. the constant function $1$ is an element of $\mathcal{H}$
  3. if $\{f_n\}$ is a sequence of non-negative functions in $\mathcal{H}$ such that $f_n\uparrow f$, where $f$ is a bounded function on $S$, then $f\in \mathcal{H}$.

Suppose futher that $\mathcal{H}$ contains the indicator of every set in some $\pi-$system $\mathcal{C}$. Then $\mathcal{H}$ contains every bounded $\sigma(\mathcal{C})-$measurable function on $S$.

So the question is: Is this the point where we have to use a monotone class argument to ensure that $R_\lambda$ is a map on $b\mathcal{E}$? And if so, how to choose $\mathcal{H}$?

share|improve this question
    
And... what is the question? –  Did Mar 24 '13 at 19:06
    
@Did if my approach is correct and if so, how to choose $\mathcal{H}$? my first thought was $\mathcal{H}$ should be all bounded measurable functions, s.t. the resolvent of it is also bounded and measurable. –  math Mar 24 '13 at 20:29
    
I think that $\mathcal{H}$ should work. It is more or less straightforward that it is a monotone class. –  Bunder Mar 30 '13 at 22:00
    
@Bunder the only two things which are not clear to me are point 3 of the montone class theorem and how to choose $\mathcal{C}$? and verify it for that system? –  math Mar 30 '13 at 22:51

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