The Original Question: Demonstrate that every martingale is a local martingale.
Attempt at a Solution:
Consider the standard setup of this problem: $\mathscr{F}_t$ is the filtration that satisfies the normal conditions, $(\sigma_n)_{n\in \mathbb{N}}$ is the monotone increasing sequence of stopping times with $\text{lim}_{n \to \infty}\sigma_n = \infty$ and $\{X_t\}_{t \in [0,\infty]}$ is a martingale.
To show that $X$ is a local martingale I will show that $E[X_\infty | \mathscr{F}_t] = X_{min(t , \sigma_n)}$.
We know that $X_\infty = X_{min(\infty,\sigma_n)}=X_{\sigma_n}$ because $\sigma_n \in \mathbb{R}_+$. It then follows that:
$E[X_{\sigma_n}|\mathscr{F}_t] = X_{\sigma_n} = X_{min(t,\sigma_n)}$.
However this incorrect for any $t < \sigma_n$, and so $\exists$ $\geq 1$ mistakes in my attempted solution.
MY Question: Am I correct? If not, what's wrong with my solution, and what's some good hints to a correct solution (or just post the correct solution).