Local martingale but not martingale On wikipedia there is an example of a local martingale which is not a martingale, but I do not understand why it is a local martingale. We have the process

$
 X_t = \begin{cases}
  W_{\min(t/(1-t),T)} &amp;\text{for } 0 \le t &lt; 1,\\
  -1 &amp;\text{for } 1 \le t &lt; \infty.
 \end{cases}$ 

where $(W_t)$ is a standard Brownian motion and $T = \inf\{ t : W_t = −1 \}$. 

The expectation is

$\mathbb{E} X_t = \begin{cases}
  0 &amp;\text{for } 0 \le t &lt; 1,\\
  -1 &amp;\text{for } 1 \le t &lt; \infty.
 \end{cases}$

This expectation is clearly discontinuous. So we have that it is not a martingale.

Now we will conclude that it is a local martingale with localizing sequence $ \tau_k = \min \{ t : X_t = k \}$ if there is such $t$, otherwise $τ_k = k$.
However, I can not figure out why this is true.  

I would appreciate some help.
I also think the wikipedia example a little bit confusing. It seems the localizing sequence can not give a martingale results since when $t\in(0,1), X_t^{\tau_n}$ is again a martingale with mean 0 and when $t\geq 1, $ it has mean -1. I have checked the reference book in that wikipedia page, and the book gives a different localizing sequence. 

Details have been discussed in this thread, and I think it is quite clear.

How to show the following process is a local martingale but not a martingale?
wrt your general question: A local martingale is not necessarily integrable, whereas a martingale is (by definition). 
This all boils down to the statement that a limit of a sequence of integrable random variables need not be integrable!