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Assume $f(t,\omega)$ is (i)separable, (ii) measurable as function from $((0,T)×\Omega)$ into $R$ and (iii) is adapted to the filtration $F_t, 0<t<T$ Also $\int_0^Tf^2(s)ds<\infty$ almost sure.
Is integral $\int_0^tf(s,\omega)ds$ measurable with respect to $F_t$?
Definition: A stochastic process $(X(t), t\in I)$ is called separable if there exists a countable sequence $(t_j)$ that is a dense subset of $I$ and a subset N of $\Omega$ with $P(N)=0$ such that, if $\omega \notin N$, then

$$\{X(t, w) \in F\ for\ all\ t\in J\} = \{X(t_j, w) \in F\ for\ all\ t_j\in J\}$$ for any open subset J of I and for any closed subset F of R
In the book they wrote:
Since the integrand is a separable process that is $F_t$ measurable, the integral is also $F_t$ measurable.
It was a part of the theorem.

Assume $f(t,\omega)$ is (i)separable, (ii) measurable as function from $((0,T)×\Omega)$ into $R$ and (iii) is adapted to the filtration $F_t, 0<t<T$ Also $\int_0^Tf^2(s)ds<\infty$ almost sure.
Is integral $\int_0^tf(s,\omega)ds$ measurable with respect to $F_t$?
Definition: A stochastic process $(X(t), t\in I)$ is called separable if there exists a countable sequence $(t_j)$ that is a dense subset of $I$ and a subset N of $\Omega$ with $P(N)=0$ such that, if $\omega \notin N$, then

$$\{X(t, w) \in F \text{ for all } t\in J\} = \{X(t_j, w) \in F \text{ for all } t_j\in J\}$$ for any open subset J of I and for any closed subset F of R
In the book they wrote:
Since the integrand is a separable process that is $F_t$ measurable, the integral is also $F_t$ measurable.
It was a part of the theorem.

Suppose $f(t,\omega)$ is separable stochastic process which is adapted to the filtration $F_t, 0<t<T$, is integral $\int_0^tf(s,\omega)ds$ measurable with respect to $F_t$? We assume that integral exists almost sure. Definition: A stochastic process $(X(t), t\in I)$ is called separable if there exists a countable sequence $(t_j)$ that is a dense subset of $I$ and a subset N of $\Omega$ with $P(N)=0$ such that, if $\omega \notin N$, then

$$\{X(t, w) \in F\ for\ all\ t\in J\} = \{X(t_j, w) \in F\ for\ all\ t_j\in J\}$$ for any open subset J of I and for any closed subset F of R

Assume $f(t,\omega)$ is (i)separable, (ii) measurable as function from $((0,T)×\Omega)$ into $R$ and (iii) is adapted to the filtration $F_t, 0<t<T$ Also $\int_0^Tf^2(s)ds<\infty$ almost sure.
Is integral $\int_0^tf(s,\omega)ds$ measurable with respect to $F_t$?  
Definition: A stochastic process $(X(t), t\in I)$ is called separable if there exists a countable sequence $(t_j)$ that is a dense subset of $I$ and a subset N of $\Omega$ with $P(N)=0$ such that, if $\omega \notin N$, then

$$\{X(t, w) \in F\ for\ all\ t\in J\} = \{X(t_j, w) \in F\ for\ all\ t_j\in J\}$$ for any open subset J of I and for any closed subset F of R
In the book they wrote:
Since the integrand is a separable process that is $F_t$ measurable, the integral is also $F_t$ measurable.
It was a part of the theorem.

Suppose $f(t,\omega)$ is separable stochastic process which is adapted to the filtration $F_t, 0<t<T$, is integral $\int_0^tf(s,\omega)ds$ measurable with respect to $F_t$? We assume that integral exists almost sure.

Suppose $f(t,\omega)$ is separable stochastic process which is adapted to the filtration $F_t, 0<t<T$, is integral $\int_0^tf(s,\omega)ds$ measurable with respect to $F_t$? We assume that integral exists almost sure. Definition: A stochastic process $(X(t), t\in I)$ is called separable if there exists a countable sequence $(t_j)$ that is a dense subset of $I$ and a subset N of $\Omega$ with $P(N)=0$ such that, if $\omega \notin N$, then

$$\{X(t, w) \in F\ for\ all\ t\in J\} = \{X(t_j, w) \in F\ for\ all\ t_j\in J\}$$ for any open subset J of I and for any closed subset F of R