I'm showing that $ \displaystyle \int_{0}^{r} B_{t} \,dt $ is a Gaussian random variable (here, $B_{t}$ is a brownian motion). I would like to know if there is any fault in the following argument:
Let $A_{r}(\omega)=\displaystyle \int_{0}^{r} B_{t}(\omega) \,dt$. Since $ B_{t} (\omega) $ is a continuous function on $ t $, by the Mean Value Theorem for integrals we would have $A_{r}=rB_{\lambda(r)}$, where $0<\lambda(r)<r$. Since $aB_{t}$ has Normal distribution, the conclusion follows.
I don't know if the continuity argument is well applied in the case of stochastic processes. I have already seen other solutions, I just want to know what fails here. Thanks
Not quite. For fixed $\omega$, you know that (a.s.) $B_t(\omega)$ is continuous. So, after you fix $\omega$, then, by the Mean Value Theorem, there exists some $\lambda_r(\omega)$ such that $$\int_0^r{B_t(\omega)\,dt}=rB_{\lambda_r(\omega)}(\omega)$$ where $0<\lambda_r(\omega)<r$. But that $\lambda_r$ might (and will) change with $\omega$; you have no control over the distribution of $\lambda_r$ as a random variable. In particular, it might have some strong correlation with $\{B_t(\omega)\}_t$. So you cannot conclude that $B_{\lambda_r(\omega)}$ is normally distributed, nor that $rB_{\lambda_r(\omega)}$ is.