Prove that the stochastic process can not have continuous paths. This problem is about stochastic processes, but what I really need help with is using the dominated convergence theorem in the end:
What I need to prove is that a stochastic process having these properties can not have continuous paths:

  
*
  
*$t_1\ne t_2 \rightarrow W_{t_1}, W_{t_2}$ independent.
  
*The process is stationary.
  
*$E[W_t]= 0, \forall t$

The hint for the exercise is: 

Consider $E[(W_t^{(N)}-W_s^{(N)})^2]$ where:
$$W_t^{(N)}=(-N)\vee(N\wedge W_t)$$

From what I understand, this means that the modified process is $-N$ if the original process is smaller than $-N$, it is equal to the original process, if the original process is between $-N$ and $N$, and it is $N$ if the original process is bigger than $N$. Basically we bound the absolute value.
Now I will try to solve the exercise by contradiction.
We get $E[(W_t^{(N)}-W_s^{(N)})^2]=2E[(W_t^{(N)})^2]$, since the expected value of each is 0, and they are independent. But $W_t^{(N)}$ is 0 iff  $W_t$ is 0, so $E[(W_t^{(N)})^2]$  can not be 0, because then $W_t$
 would be 0 a.s. and then we would only have a trivial process.
Notice that stationarity gives that $E[(W_t^{(N)}-W_s^{N})^2]=2E[(W_t^{(N)})^2]$ must be constant no matter what s is.
Now look at $E[(W_t^{(N)}-W_{t+1/N}^{(N)})^2]$ for $\omega$ fixed we will get N is bigger than $W_t(\omega)$ at some $N$, assuming for contradiction continuity of the path, it will happen in a neighbourhood around $W_t(\omega)$ the neighbourhood is if we hold $\omega$ fixed, it is a neighborhood in the time-parameter. Then the assumed continuity gives that we get a pointwise limit to $0$.
Now, if I could move the limit outside the integral. I would get a contradiction, because then we would must have that $E[(W_t^{(N)}-W_{t+1/N}^N)^2]\rightarrow 0$. But we showed above that this is a constant value, which has to be above $0$, or else the process is trivial.
Can you please help me finish the proof? I need a dominating function to move the limit outside the integral. But I don't see how the $N$ boundary helps us, because $N$ itself grows without bound? Any tips?
 A: There is not even a measurable process satisfying your conditions 1., 2., 3.
That is, suppose that $\Omega\times[0,1]\ni(t,\omega)\mapsto W_t(\omega)$ is $\mathcal F\otimes\mathcal B([0,1])$-measurable on some probablity space $(\Omega,\mathcal F,\Bbb P)$, and satisfies your conditions 1., 2., and 3. In addition, truncate as you do, if necessary, to reduce to the case in which $|W_t(\omega)|\le C$ for some constant $C\in(0,\infty)$. Finally, eliminate the trivial case by assuming that $\Bbb E[W_t^2]>0$.
Consider now the integral $I_T(\omega):=\int_0^T W_t(\omega)\,dt$, for $0<T\le 1$, which is a random variable because of the measurability assumption. By Fubini,
$$
\Bbb E[I_T^2]=\int_0^T\int_0^T\Bbb E[W_sW_t]\,ds\,dt =0,
$$
because $\Bbb E[W_sW_t]=0$ by 1. and 3. Thus $I_T=0$, a.s., for each $T$.
But $T\mapsto I_T(\omega)$ is an absolutely continuous function for each $\omega$, so differentiating with respect to $T$ and invoking Fubini and condition 2., we get $W_t=0$ a.s., for each $t$, in violation of the non-triviality assumption.
