Is continuous L2 bounded local martingale a true martingale? I can prove it briefly, but I found a "counter" example. (There must be a mistake in the following words...)
I can prove: X is a continuous local martingale, with $X_0=0$ a.s, then X is $L_2$ bounded iff $E[X]_\infty<\infty$
I can prove: X is a continuous local martingale, if $E[X]_\infty<\infty$, then X is a true martingale.
It seems that we can get: any $L_2$ bounded continuous local martingale is a true martingale.
But there is a "counter" example:
If X is a standard Brownian motion in $R^3$, started at $0$, 
I can prove
(i) $Y_t =1/|X_{1+t}|$ is a local martingale;
(ii) Y is bounded in $L^2$;
(iii) Y is not a martingale.
Can I say: since Y isn't continuous, therefore Y is bounded in $L^2$, but not a martingale??? But I guess that we can prove Y is continuous almost surely.
Is continuous L2 bounded local martingale a true martingale???
 A: No, not every continuous $L^2$ bounded local martingale is a true martingale (see your counterexample). However, every local martingale with $L^1$ bounded quadratic variation is a true martingale!
The problem is in your second statement:
If $X$ is a continuous martingale with $L^2$-bounded quadratic variation: $\mathbb{E}[X]_\infty <\infty$ THEN $X_t$ is $L_2$ bounded. The converse ('if f ') is not true in general (which is what is going wrong in your example).
Let $\tau_n$ be a localizing sequence. we then have:
$$
\mathbb{E}X_t^2 =  \mathbb{E}\lim_{n \rightarrow \infty} X_{t \wedge \tau_n}^2 \leq \lim_{n \rightarrow \infty} \mathbb{E} X_{t \wedge \tau_n}^2 = \lim_{n \rightarrow \infty} \mathbb{E} [X]_{t \wedge \tau_n}= \mathbb{E} \lim_{n \rightarrow \infty} [X]_{t \wedge \tau_n} =E[X]_t
$$
The inequality follows from Fatous lemma. The second exchange of limit and integration is, however, justified by the monotone convergence theorem. 
In your specific example, Fatou's lemma results in a strict inequality:
$$
\frac{1}{1+t} = \mathbb{E} Y_t^2 < \mathbb{E}[Y]_t = \mathbb{E} \int_0^t \frac{1}{X_{1+t}^4} ds =  \int_0^t \mathbb{E}Y_s^4 ds = \infty
$$
The exchange of integrators is justified because $Y_s^4$ is positive. Further, while the second moment of $Y_s$ exists, the fourth moment does not. Hence the last value is infinity.
