Ito integral is a martingale

If $$\{g(t)\},$$ is deterministic, then we can use the fact that, the Ito-integral is normally distributed. Hence, using the properties of an Ito-integral on slide set 3, slide 13, we have $$X_t = \int_{0}^{t} g_s dW_s,$$, this means that there is no drift (easiest way to check is the fact that there is no dt term) and we can conclude that its expectation is 0. Moreover, if we can prove that $$\mathbb{E} [M_t] = M_0$$ (in this case we would have X(0) = 0), this will indicate that the Ito-integral has no drift process and hence meets the definition of a martingale (provided it meets certain criteria that needs to be made explicit).