Question regarding Brownian motion Hello I have two questions regarding the construction of Ito's integral in Øksendals book from here:
http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf
On page 25 he lists these 3 properties that has to hold for the integrand: 


*

*$\mathcal{B}\times \mathcal{F}$-measurable

*$\mathcal{F}_t$-adapted

*$E [\int_S^T f^2 dt]< \infty$
$\mathcal{F}_t$ is the natural filtration for the brownian motion.
But after he has finished construction the integral he integrated with respect to a brownian motion in an example. But how do we know that the first and third property really hold for a Brownian motion? We know that each $B_t$ is $\mathcal{F}$-measurable. Buw why is it joint measurable? And the last integrability requirement? I am not sure how to show that? I know that if we keep $\omega$-fixed we are ok, because we have continuity over a closed and bounded interval, but how do we not know that when taking expectation it blows up? I don't see that we have some kind of uniform boundedness with respect to $\omega$
For the first property I get that if $O$ is an open set, than all $B(s,\omega)^{-1}(O)=\cup_{g \in \mathbb{Q}}B_q^{-1}(O)$, because of the continuity of the brownian motion. But this is a subset of $\mathcal{F}$ not $\mathcal{B} \times \mathcal{F}$. Any hints on how to get this to work? The farthest I get is that the set $\{(q,\omega): B(q,\omega)\ \in O, q \in \mathbb{Q}\}$ is mesurable. But I don't know how to extend since the irrationals are uncountable, and the sigma-algebra is only closed under countable operations.
For the second problem I have no more ideas of what to try.
 A: Let $(B_t)$ be a Brownian motion (BM) on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and let $(\mathcal{F}_t)$ denote the natural filtration of BM. As you noticed BM as a stochastic process is adapted to its natural filtration, and so point 2. is satisfied. 
Another property from the definition of BM is the continuity of its paths, that is, for any $\omega \in \Omega$ the path $t \mapsto B_t(\omega)$ is continuous. 
This gives us point 1, which would be satisfied even if we assume a weaker condition, that is, if our stochastic process has right (left)-continuous paths.  
Theorem.
If a stochastic process $(X_t)$ such that $X_t \colon \Omega \rightarrow \mathbb{R}^n$ has right-continuous paths, then it is $\mathcal{B}(\mathbb{R}_+) \otimes \mathcal{F}$-measurable. 
Proof.
For $n\in \mathbb{N}$, $t \geq 0$, $\omega \in \Omega$ we define
$$X_t^n(\omega) := \sum_{k\geq 1}\mathbf{1}_{ \left[ \frac{k-1}{2^n} , \frac{k}{2^{n}}\right)}(t) X_{\frac{k}{2^n}}(\omega).$$
Let $E \in \mathcal{B}(\mathbb{R}^n)$, then for all $n \in \mathbb{N}$ we have
$$ (X_t^n)^{-1}(E)= \{ (t, \omega) \colon X_t^n(\omega) \in E\} = \bigcup_{k\geq 1} \left[\frac{k-1}{2^n}, \frac{k}{2^n} \right) \times \{\omega \colon X_{\frac{k}{2^n}}(\omega) \in E \} \in \mathcal{B}(\mathbb{R}_+) \otimes \mathcal{F}.$$
Hence, for each $n \in \mathbb{N}$, $(X_t^n)$ is $\mathcal{B}(\mathbb{R}_+) \otimes \mathcal{F}$-measurable.
It is sufficient to show that $X^n \rightarrow X$ pointwise on $\mathbb{R}_+ \times \Omega$ as $n \to \infty$, because the pointwise limit of measurable functions is measurable (link). 
For each $n \in \mathbb{N}$ there exists a unique $k_n$ such that $t \in \left[\frac{k_n-1}{2^{n}}, \frac{k_n}{2^n} \right)$ and $X_t(\omega) = X_{\frac{k_n}{2^n}}(\omega)$. Since $\frac{k_n}{2^n} \downarrow t$ as $n\to \infty$, right-continuity yields that 
$$X_t^n(\omega) = X_{\frac{k_n}{2^n}}(\omega) \stackrel{n \to \infty}{\longrightarrow} X_t(\omega). \qquad  \square$$
Same holds if $(X_t)$ left-continuous paths, change $[a,b)$ to $(a,b]$ in the definition of $X_t^n$.
To show that point 3. note that
$$ \mathbb{E}\left[\int_S^T B_r^2 \ \mathrm{d}r\right] = \int_S^T \mathbb{E}[B_r^2] \ \mathrm{d}r = \int_S^T r \ \mathrm{d}r = \frac{1}{2}\left(T^2-S^2\right)< \infty .$$
