Let $(W_u)_{u\geq0}$ be a Wiener process defined on a probability space $(\Omega,\mathbb P, \mathscr{F}, \sigma(W_u)_{u\geq0})$ where $\sigma(W_t)$ is the $\sigma$-algebra generated by $W_t$. Let $s>t$.
Is $W_t$ mesurable with respect to $\sigma(W_s)$? My understanding is that it is not the case, but I am unsure on how to prove it.
Background: I have read that for a Brownian Motion $(W_u)_u$, if $0<t<s$ then: $$E\left(W_tW_s|W_s\right)=\frac{t}{s} W_s^2$$ whereas I would have expected it to be equal to $W_tW_s$. The reason given for the result is that $W_t$ is not $\sigma(W_s)$-measurable, i.e. $E(W_t|W_s)\not= W_t$. I am trying to understand this.