Let $B = (B_t)_{t \in [0,T} $ a Brownian motion. Let $H = (H_t)_{t \in [0,T}, G = (G_t)_{t \in [0,T} $ be progressively measurable.
What are the conditions on $H,G$ such that $(\int_0^t H_s G_s dB_s)_{t \in [0,T]} $ is a true martingale?
From what I know we should have $$ E \left[ \int_0^T \mid H_s G_s \mid^2 \; ds \right] < \infty. $$
This in turn would be guaranteed if $$ E \left[ \int_0^T |H_s|^4 + |G_s|^4 \; ds \right] < \infty. $$
Is it possible to have a weaker condition to get a martingale?
Edit: Is it enough to ask for $$ \int_0^T | H_s G_s |^2 \; ds < \infty, \quad P-a.s. ? $$
Edit: I was missing some information. $G$ is $dt\otimes P$ square integrable and $E[ \sup_{t \in [0,T] } |H_s|^2 ] < \infty. $ Then $$ E\left[ \left(\int_0^T | H_s G_s |^2 \; ds \right)^{\frac{1}{2}} \right] < \infty. $$
Thus, $ \int_0^\cdot H_s G_s ds $ is a true martingale by BDG-inequality.