Stochastic integral of $L^2$ function is martingale Let $f: \mathbb{R} \to \mathbb{R}$ a function with $f \in L^2_{[0,1]}$ for each $t \ge 0$, $(B_t)_t$ a Brownian motion process on a probability space $\Omega$ and $(\mathcal{F}_t)_t$ the canonical filtration with respect to $(B_t)_t$.
I want to show that $M_t := \int_0 ^t f(s) dB_s$ is a martingale.
The main two problems are how to see that :
-$(M_t)_t$ is adapted with pespect to filtration $(\mathcal{F}_t)_t$
-and the most crucial point, the fact that for $r \le t$ we have the martingale property for conditional expectation, i.e. :
$$\mathbb{E}(M_t \vert \mathcal{F}_r) =\mathbb{E}(\int_0 ^t f(s) dB_s \vert \mathcal{F}_r) = \int_0 ^r f(s) dB_s = M_r$$
 A: I think that you can construct the proof starting from step functions and then develop for all generic functions. The main point is the deterministic property of function $f$.
We consider a step function $f$ of kind $\sum_{=1}^n a_i\mathbb{1}_{[t_i,t_{i+1})}$ . By the definition of stochastic integral we obtain:
$$\int_0^rf(s)dB_s=\sum_{i=1}^na_i(B_{r\wedge t_{i+1}}-B_{r\wedge t_i}) $$
And in this case it is of course a martingale (we use the independence property of Brownian increments).
Every $L^2$ functions can be approximated by step functions and then you can complete the prof.
The stochastic integral of generic random variable is defined as $L^2$-limit (it can be defined for a bigger class of function but in this case we stay in the special class of deterministic function, so we use the $L^2$ limits).
Consider a sequence of step functions $\{f^n\}_n$ that converges to $f$.
From previous step we have:
$$E(\int_0^tf^n(s)dB_s|\mathcal{F}_r)=\int_0^rf^n(s)dB_s\quad\forall n$$
and for costruction that I have said before:
$$\int_0^rf^n(s)dB_s\quad\to^{L^2} \int_0^rf(s)dB_s $$
Now we use the following results.
If $X_n\to^{L^2}X$ then, conditioned on any $\sigma$-algebra, it holds:
$$E(X_n|G)\to^{L^2} E(X|G)$$
Then we have complete.
Last result can be showed by Jensen inequality.
