Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(\Omega, \mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P})$ be a probability space.

Let $C$ be the stochastic processes which can be written on the form $\sum _{i=1}^n K_i 1_{(a_i,b_i]}$ for $a_i,b_i \in \mathbb{R}$ and $K_i$ measurable wrt $\mathcal{F}_{a_i}$.

Let $\mathcal{P}=\sigma\{X: \text{X left cont. adapted, bounded}\}$.

One can then show that $\sigma(\mathcal{C})=\mathcal{P}$.

Let $$\mathcal{H}=\{h(s) \in \mathcal{P} \;:\; \exists (h_n(s))_{n\geq 1} \subset \mathcal{C} \text{ s.t. } \lim_{n\to\infty}E [\int_0^\infty (h_n(s)-h(s))^2]=0 \}$$

Then $\mathcal{H}$ is a vectorspace. But I would like it to be closed with respect to bounded monotone convergence to apply a monotone class argument. That is if $h_m\in \mathcal{H}$ for all $m$, $|h_m|<K$ and $h_m(s)\to h(s)$ monotonely then $h(s)\in \mathcal{H}$. My lecture stated it is, but how can I realize it?

share|cite|improve this question

Let $h_n(s)\in \mathcal{H}$ $\forall n$ s.t. $0\leq h_n(s) \uparrow h_s$. For each $n$ we can find $h_{n,k}(s)$ s.t. $\lim_{k\to \infty} E\int_0^\infty h_{n,k}(s)-h_n(s) = 0$ in particular we can find $\{k_n\}_{n\geq 1}$ s.t $ E\int_0^\infty h_{n,k_n}(s)-h(s) \leq n^{-2}$. Now it follows that $\lim_{n \to \infty} E\int_0^\infty h_{n,k_n}(s)-h(s) =0$ as wanted.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.