Show that the stochastic exponential is a true martingale Let $W = \{W_t : t\ge0\}$ be a standard Brownian motion on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $f$ be a deterministic function such that 
$$
\int_0^tf^2(s)\,ds<\infty
$$
for all $t\ge 0$. Show that the stochastic exponential 
$$
M_t = \exp\left(\int_0^tf(s)\,dW_s - \frac{1}{2}\int_0^tf^2(s)\,ds\right)
$$
is a martingale. 
I'm aware that this can be verified immediately using Novikov's criterion, for example, but I'm looking for a more direct proof than this. It's easy to demonstrate using Ito's lemma that 
$$
M_t = 1 + \int_0^tf(s)M_s\,dW_s,
$$
and so the result can be boiled down to showing that 
$$
\mathbb{E}\left[\int_0^tf^2(s)M_s^2\,ds\right]<\infty.
$$
This seems promising, but I'm unsure how to proceed. I should mention that the subsequent part of the exercise I'm trying to solve asks to use the martingale property of $M_t$ to show that 
$$
\int_0^tf(s)dW_s\sim N\left(0,\int_0^tf^2(s)\,ds\right).
$$
This is straightforward to do, but it suggests that the martingale property can be demonstrated without appealing to these facts. Any suggestions or references would be greatly appreciated. 
 A: Itô's formula shows that
$$M_t = 1+ \int_0^t f(s) M_s \, dW_s \tag{1}$$
and this implies, in particular, that $(M_t)_{t \geq 0}$ is a local martingale. (Note that $(M_t)_{t \geq 0}$ has continuous sample paths, and therefore the stochastic integral on the right-hand side is well-defined.) On the other hand, it follows from the very definition that $M_t \geq 0$ for each $t \geq 0$. Since any non-negative local martingale is a supermartingale (see e.g. this question for details), we conclude that $(M_t)_{t \geq 0}$ is a supermartingale. Thus,
$$\mathbb{E}(M_t) \leq \mathbb{E}(M_0) \leq 1$$
which implies
$$\mathbb{E} \exp \left( \int_0^t f(s) \, dW_s \right) \leq \exp \left( \frac{1}{2} \int_0^t f(s)^2 \, ds \right)$$
for each $t \geq 0$. Replacing $f$ by $2f$ we find in particular that
$$\mathbb{E} \left| \exp \left( \int_0^t f(s) \, dW_s \right) \right|^2 = \mathbb{E}\exp \left( \int_0^t 2f(s) \, dW_s \right) \leq \exp \left(  2 \int_0^t f(s)^2 \, ds \right) < \infty,$$
and so
$$\mathbb{E}(M_t^2) \leq \exp \left( \int_0^t f(s)^2 \, ds \right).$$
Using this estimate and the fact that $f$ is deterministic, we can easily check that the stochastic integral $\int_0^t f(s) M(s) \, dW_s$ is a true martingale, and now $(1)$ shows that $(M_t)_{t \geq 0}$ is a martingale.
