Consider the following $\mathcal F_t$- (continouous) local martingale $$M_t = \int_0 ^t \exp{((B_2(s)^2)} dB_1(s)$$ where $\left(B_t\right)_{t\geq0} =\left(B_1(t),B_2(t)\right)_{t\geq0}$ is $\mathcal F_t$-Brownian motion in $\mathbb R^2$, null at $t=0$.
How to show that $\left(M_t \right)_{t\geq 0}\notin \mathcal{M }^2_c=\{\mathcal {F}_t - $real continuous martingale square integrable, with $ M_0 =0 \}$ ?
Let $\mathcal G_t=\sigma(B_2(s),s\leqslant t)$. Since $(B_1(s))_{s\leqslant t}$ is independent of $\mathcal G_t$ and $(B_2(s))_{s\leqslant t}$ is measurable with respect to $\mathcal G_t$, Itô's isometry indicates that $$ \mathbb E(M_t^2\mid\mathcal G_t)=\int_0^t\mathrm e^{2B_2(s)^2}\,\mathrm ds. $$ Everything is nonnegative, hence $$ \mathbb E(M_t^2)=\int_0^t\mathbb E(\mathrm e^{2B_2(s)^2})\,\mathrm ds. $$ Since $B(s)$ is centered normal with variance $s$, $\mathbb E(\mathrm e^{2B_2(s)^2})$ is infinite for every $s\geqslant\frac14$. Thus, $\mathbb E(M_t^2)$ is infinite for every $t\gt\frac14$ and in particular, $(M_t)_{t\geqslant0}$ is not a square integrable martingale.
(Note that $\mathbb E(M_t^2)=\frac12(1-\sqrt{1-4t})$ for every $t\leqslant\frac14$, hence $(M_t)_{t\leqslant1/4}$ is a square integrable martingale.)