If we want $\int_0^⋅XdM$ to be a martingale, do we need to assume $E[\int_0^tX_sd[M]_s]<∞$ for all $t$ or even $E[\int_0^∞X_sd[M]_s]<∞$?

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
• $(\mathcal F_t)_{t\ge0}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$
• $(M_t)_{t\ge0}$ be a real-valued continous square-integrable $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$
• $(X_t)_{t\ge0}$ be a real-valued $\mathcal F$-predictable process on $(\Omega,\mathcal A,\operatorname P)$
• $T\in[0,\infty]$

In the construction of the Itō integral process $$X\cdot M=\left(\int_0^tX_s\:{\rm d}M_s\right)_{t\ge0}$$ as a square-integrable martingale, do we need to assume $$\operatorname E\left[\int_0^t|X_s|^2\:{\rm d}[M]_s\right]<\infty\;\;\;\text{for all }t\in[0,T]\setminus\left\{\infty\right\}\tag1$$ or even $$\operatorname E\left[\int_0^T|X_s|^2\:{\rm d}[M]_s\right]<\infty\tag2?$$

Clearly, the question is trivial if $T<\infty$ and the whole construction is absolutely clear to me in that case. But what if $T=\infty$? Is there any reason why we should even need $(2)$?