Certain transformation of Brownian motion is a submartingale

I have a question about a proof in Protter. Let $B$ Brownian motion and $u$ a harmonic (subharmonic) function. Then $u(B)$ is a local martingale (submartingale). I was able to show the case of local martingale by myself. The case for subharmonic functino should be similar as Protter suggest. We have by Itô

$$u(B_t)=u(B_0)+\int_0^t\nabla u(B_s)dB_S+\frac{1}{2}\int_0^t \Delta u(B_s)ds$$

Using subharmonicity of $u$ we have:

$$u(B_t)\ge u(B_0)+\int_0^t\nabla u(B_s)dB_S$$

How does this imply the submartingale property of $u(B)$?

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Fix $a<b$. After having applied Ito's formula, we get $$u(B_b)-u(B_a)=\int_a^b\nabla u(B_s)dB_s+\frac 12\int_a^b\Delta u(B_s)ds\geq \int_a^b\nabla u(B_s)dB_s.$$ Therefore, writing the definition of stochastic integral and taking the condition expectation with respect to $\sigma(B_a)$, we can see that the conditional expectation of the integral is $0$. Indeed, it's the limit of a sum of terms of the form $E[\nabla(B_{t_i})(B_{t_{i+1}}-B_{t_i})\mid B_a]$. As $\nabla u(B_{t_i})$ is independent of $B_a$, we get the wanted result.
Two questions: First, the resulting inequality would be $E[u(B_t)]\ge u(B_0)$. I believe you can't use this to show $E[u(B_t)|\mathcal{F}_s]\ge u(B_s)$. Moreover why should the integral vanish? A priori it is a local martingale, if $\nabla u$ is bounded, then it is a true martingale for which your claim would be true. For a local martingale, you have to stopped first using a localizing sequence. – user20869 Sep 24 '12 at 11:13