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I know that $\int_0^{\infty}e^{-\alpha t}c(X_t)dt$ is a random variable when $c(.)$ is a measurable function and $X_t$ is a stochastic process. How can this be proved rigorously?

bc78
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    Note that you have to ensure that the integral exists, i.e. $e^{-\alpha t} c(X_t) \in L^1(dt)$. And do you denote by $\int_0^{\infty} \dots ,dt $ an (improper) Riemann integral or a Lebesgue integral? – saz Jun 24 '15 at 11:44
  • It is an improper Riemann integral. What I want to see is that improper Riemann integral of a random variable is a random variable. Thank you very much for your clarification since I forgot to write that c is also bounded. – bc78 Jun 24 '15 at 11:49
  • @Bora I am guessing the interpretation of the integral is as follows. You fix $\omega$ and integrate $X_{\omega}(t)$. If that is the case, then shouldn't you first say something about the paths of $X(t,\omega)$ for the integral to exist in the Riemannian sense? – Calculon Jun 24 '15 at 12:08
  • Thank you very much @Calculon, I have missed that part. $X_t$ is a stable Markov process. That is why $X_t(w)$ has at most countably many discontinuities. Moreover, in my previous comment to saz, I have said that $c$ is also a bounded function. I guess that is why Riemann integral should exist. – bc78 Jun 24 '15 at 14:29
  • @Bora I haven't really done anything but you are welcome. I was also thinking whether $c(\cdot)$ being bounded would force $c(X_\omega(t))$ to have sufficient path regularity for your integral. So far I don't know the answer to that. – Calculon Jun 24 '15 at 14:32
  • @Bora No, this doesn't ensure that $t \mapsto c(X_t)$ is Riemann-integrable. For example $c(x) := 1_{\mathbb{Q} \cap [0,1]}(x)$ is a bounded (Borel)measurable function but it is not Riemann-integrable (in particular, $c(X_t)$ is in general not Riemann-integrabe for a cadlag process $X$). So you need some additional assumptions .... – saz Jun 24 '15 at 14:55
  • @saz ok you are right. I would like to assume that $X_t$ is a regular Markov process which has right continuous sample paths. That is why it has at most countably many discontinuities. I think this ensures that the Riemann integral exists. What I could not do is to show that this integral is a measurable function. I want to do this from scratch by using the definition of measurability or the definition of random variables. I know that $\sup$ and $\inf $ are measurable functions. I guess these can be used to complete the proof but I could not do it in a way that satisfies me. – bc78 Jun 25 '15 at 05:01
  • @Bora No, it is not enough to assume that $(X_t)$ is a regular (cadlag) MP. As I said, we need additional assumptions to guarantee that $t \mapsto c(X_t)$ is Riemann integrable and right-continuity of $(X_t)_t$ does not imply this. – saz Jun 25 '15 at 05:12
  • @saz In the following link, you will be able to see that a function having at most countable discontinuities is Riemann integrable:

    http://math.stackexchange.com/questions/263189/proof-that-a-function-with-a-countable-set-of-discontinuities-is-riemann-integra

    In the following link (Chapter 3), the proposition says that a right continuous function has at most countable discontinuities:

    http://www.math.wisc.edu/~kurtz/735/main735.pdf

    So, where is the problem?

     – bc78 Jun 25 '15 at 08:58
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    @Bora Well, you assume that $X$ has at most countable many discontinuities, but (at least if I didn't miss it) you don't assume anything on $c$. just consider $X_t := t$ (which hasn't discontinuities at all) and $c(x) :=1_{\mathbb{Q} \cap [0,1]}(x)$, then $c(X_t)$ is not Riemann integrable. – saz Jun 25 '15 at 09:12
  • @saz Now, I have got the point I guess. What I assume is that $X$ is a regular Markov process, which is a continuous time process with a countable state space. Since its paths are right continuous and have at most countable many discontinuities, the general structure of its sample paths is a kind of step function with countably many jumps. That is why, I think, boundedness of $c(.)$ should be sufficient. – bc78 Jun 25 '15 at 10:54

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