Argument that vector space is closed with respect to bounded monotone convergence 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? 

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.
