# Square Integrable Martingales and the Unit Process

Let $X_t$ be a continuous square-integrable martingale. Then it is true (I think, please correct me if I am wrong) that:

$$\forall t \in [0,\infty), \quad \int_0^t 1_{[0,t]}(s) dX_s = X_t - X_0$$

So couldn't we use this as a definition of $\int_0^t 1 (s) dX_s$ (where $1$ is the constant function)? I.e.

$$\forall t \in [0,\infty), \quad \int_0^t 1(s) dX_s = \int_0^t 1_{[0,t]}(s)dX_s$$

I am confused because the integral on the right is supposed to be a square integrable martingale, but the one on the left is supposed to be just a local martingale, because the integrand on the left is not square-integrable, only locally $L^2-$bounded, whereas the integrand on the right is square-integrable.

Where is my mistake?

square-integrable = $L^2$-bounded: $\mathbb{E}[\int_0^{\infty} |f_s|^2 \mathrm{d}s]< \infty$

locally square-integrable/$L^2$-bounded: $\forall t \in [0,\infty) \quad \mathbb{E}[\int_0^t |f_s|^2 \mathrm{d}s] < \infty$

Seemingly we could do the same trick for any locally $L^2-$bounded process, and we would always get a square integrable martingale for any finite $t$ (just integrate $(f\cdot 1_{[0,t]})(s)$ instead of $f(s)$).

But Revuz and Yor says that the integrand needs to be square integrable, not just locally square integrable, for the stochastic integral to again be a continuous square-integrable martingale, and that if the integrand is only locally square integrable, then the resulting process is a continuous local martingale, but not a true martingale.

• You seem to be confusing time intervals and state intervals. For example, the meaning of $$\int_0^t 1_{[0,t]} dX_t$$ is quite unclear since $1_{[0,t]}$ is defined on $\mathbb R_+$ while every $X_s$ is defined on $\Omega$. On the other hand $$\int_0^\infty 1_{[0,t]}(s)dX_s=\int_0^tdX_s=X_t-X_0.$$ – Did Jul 19 '16 at 8:43

Actually re-reading the section of Revuz and Yor which I cited earlier, that appears to be what they are discussing before moving on to discuss local martingale integrals. My professor told me once (seemingly in error or more likely I wasn't paying attention and misheard them) that Brownian motion is a local martingale but not a true martingale, and now I have difficulty convincing myself that this is not true. It doesn't matter that Brownian motion isn't integrable, since it is integrable on $[0,t]$ for all $t$, that is enough.