Showing Ito's formula gives a semimartingale decomposition for brownian motion

I'm trying to show that for a standard Brownian motion and some twice continuously differentiable function $$f$$ that $$f(B_t)$$ is a local martingale iff $$f'' = 0$$.

Applying Ito's formula gives, and since $$\langle B_s\rangle = s$$: $$f(B_t) = f(B_0) + \int_0^t f'(B_s) dB_s + \frac{1}{2} \int_0^t f''(B_s) ds$$

Ito's formula tells us that $$f(B_t)$$ is a semimartingale, so has unique decomposition $$f(B_0) + M_t + A_t$$ where $$M_t$$ is a local martingale and $$A_t$$ is a process of finite variation. Presumably we use uniqueness of the decomposition to match up the terms of these expressions and the result follows, so my only questions are why

1. $$\int_0^t f'(B_s) dB_s$$ is a local martingale?
2. $$\int_0^t f''(B_s) ds$$ is a process of finite variation?

Sorry for the long preamble, thank you for any help!

• Why not take a look at the literature? The (local) martingale property of the stochastic integral with respect to Brownian motion is discussed in (almost) any book on this topic. – saz Jun 10 at 16:35
• I think I understand why the integral w.r.t brownian motion is a martingale, and it doesn't seem like the inclusion of $f'(B_s)$ makes much difference. The integral in (2) seems like a less standard result though, what exactly is the integral of $f''(B_s)$ w.r.t. $s$ defined as? – watt Jun 10 at 17:41