Non-existence of this stochastic process? I am intrigued by the following statement in Oksendal , Stochastic differential equations, Chapter 3:

(A) 
There does not exist a 'reasonable' stochastic process satisfying the following conditions:
(i) $t_1 \ne t_2 \implies$ $W_{t_1}$ and $W_{t_2}$ are independent
(ii) $W_t$ is stationary ie the joint distribution of $\{W_{{t_1}+t}...W_{{t_k}+t}\}$ does not depend on t
(B) 
Moreover if we require $E[W_t^2]=1$ then the function $(t,\omega)\rightarrow W_t(\omega) $ cannot be measurable with respect to the $\sigma$-algebra $B \text{ x } F$ where $B$ is the Borel $\sigma$-algebra on $[0,\infty[ $ (See Kallianpur 1980 p10)


I do not have access to that reference. Is there another one that someone could suggest to explain these statements?
 A: The variance computation shows that $\int_I X_t\,dt=0$ a.s.; the null set $N_I$ just acknowledges the fact that this "almost surely" depends on the interval $I$. The grand null set $N$ then has the property that
$$
\omega\notin N \Longrightarrow \int_IX_t(\omega)\,dt=0,\quad\hbox{for all } I\hbox{ with rational endpoints}.
$$
To be extra careful, notice that the $E\left[\int_0^1 |X_t|\,dt\right]<\infty$, so there is one more null set $N'$ such that
$$
\omega\notin N'\Longrightarrow t\mapsto X_t(\omega)\in L^1([0,1]).
$$
It $\omega\notin N'$ then $\int_a^b X_t(\omega)\,dt$ is a continuous function of $a$ and $b$; it now follows that in $\omega\notin N\cup N'$ then $\int_a^b X_t(\omega)\,dt = 0$ for all real $a,b$ with $0\le a\le b\le 1$. You can now apply the Lebesgue differentiation theorm to conclude that if $\omega\notin N\cup N'$ then $X_t(\omega)= 0$ for (Lebesgue) a.e. $t\in[0,1]$. This leads, via Fubini, to a contradiction  of the assumption that $E[X_t^2]=1$ for each $t\in[0,1]$.
A: Here is the desired result, still looking for an answer worth a bounty as this answer could use additional explanation. For instance:
After the part about applying Fubini's theorem,
It says $\int_I X_t(\omega)dt = 0$ for $\omega \notin N_I$ where $P(N_I) = 0$, then they construct this union $N = \cup_I N_I$ such that $P(N) = 0$-- it is not entirely clear why the introduction of $N$.
Nonetheless here is the result

