Demonstrate that every martingale is a local martingale. 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).
 A: This is too long to be a comment, so I posted it as an answer instead. Let me first introduce the definitions I use of a martingale, local martingale and localizing sequence.

A sequence $(\sigma_n)_{n\in\mathbb{N}}$ is called a localizing sequence if $\sigma_n$ is a stopping time for all $n\in\mathbb{N}$ and $\sigma_n\uparrow \infty$ a.s.
A stochastic process $(X_t)_{t\geq 0}$ is called a martingale (wrt. $(\mathscr{F}_t)_{t\geq 0}$) if it is integrable, adapted and satisfies
  $$
E[X_t\mid\mathscr{F}_s]=X_s \quad \text{a.s.}
$$
  for every $0\leq s<t$.
A stochastic process $(X_t)_{t\geq 0}$ is called a local martingale if there exists a localizing sequence $(\sigma_n)_{n\in\mathbb{N}}$ such that the stopped process $(X_t^{\sigma_n})_{t\geq 0}=(X_{t\wedge \sigma_n})_{t\geq 0}$ is a martingale for every $n\in\mathbb{N}$.

Now, you're starting with a martingale $(X_t)_{t\geq 0}$ and in order to show that it is also a local martingale, you just have to find a localizing sequence $(\sigma_n)_{n\in\mathbb{N}}$ such that $(X_{t\wedge\sigma_n})_{t\geq 0}$ is a martingale. Your task is now to find such sequence using the fact that $(X_t)_{t\geq 0}$ already is a martingale. To this end, note that a stopping time is a mapping from $\Omega$ into $\mathbb{R}_+\cup \{\infty\}$ and not into $\mathbb{R}_+$.
