Consider a discrete time stochastic process $X: T \times \Omega \mapsto \mathbb{R}$ defined on a probability space $(\Omega, \mathcal{F}, \mu)$. Being defined as the smallest sub-$\sigma$-algebra of $\mathcal{F}$ such that all previous and the current random variable are measurable, I understand that the natural filtration $\mathcal{F}^X_t$ is to be interpreted as the information revealed by the process up until time $t$. See below an explanation of how I wrongfully arrive at the conclusion that $\mathcal{F}^X_t$ should instead be the smallest sub-$\sigma$-algebra of $\mathcal{F}$ such that all random variables are measurable up until time $t+1$.
Suppose $T=\{1,2\}$, $\Omega = \{\omega_a,\omega_b,\omega_c,\omega_d\} = \{(0,0), (0,1), (1,0), (1,1)\}$. Suppose further that our random variables are given by $X_1(\omega_a)$ = $X_1(\omega_b) = 0$, $X_1(\omega_c)$ = $X_1(\omega_d) = 1$ and $X_2(\omega_a)$ = $X_2(\omega_c) = 0$, $X_2(\omega_b)$ = $X_2(\omega_d) = 1$ and that we observe $X_1 = 1$ at time $t=1$. Following the (proper) definition, the natural filtration $\mathcal{F}^X_1$ ensures that $X_1$ is measurable, so that $\mathcal{F}^X_1 = \{\emptyset, \Omega, (\omega_a, \omega_b), (\omega_c, \omega_d)\}$, which, at least from my point of view, does not seem to reveal anything about the history of the path. However, if we were to instead follow my alternative definition and thus condition on what we observe today, we would have $\mathcal{F}^X_0 = \{\emptyset, \Omega, (\omega_a, \omega_b), (\omega_c, \omega_d)\}$ and, more importantly, $\mathcal{F}^X_1 = \{\emptyset, \Omega, (\omega_a, \omega_b), (\omega_c, \omega_d), \omega_c, \omega_d\}$, which indeed correctly reveals that neither $\omega_a$ nor $\omega_b$ have occurred, which in turn implies $X_1 = 1$ as required.
At which point is my intuition letting me down?