We say $(W_t)_{t\in[0,T] }$ is a Brownian Motion on $(\Omega, F, F_W, P)$. For $\lambda\in \mathbb{R}$, we define $$Q(A):=E[1_AM_T], \quad A\in F,$$ where $$M_t:=exp(-\lambda W_t - \frac{1}{2}\lambda^2t).$$
I have proven $Q$ is a prob. measure, $Q$ is equivalent to $P$ and that $(M_t)_{t\in [0,T]}$ is a $P$ martingale.
Now I want to prove that $(e^{-rt}S_t)_{t\in [0, T]}$ is a $Q$ martingale,
with $S$ (under $P$) described with: $$dS_t=(\lambda \sigma + r)S_t\text{d}t+\sigma S_t \text{d}W_t, \quad r, \sigma >0.$$
With the help of Ito's formula I got: $$\text{d}(e^{-rt}S_t) = e^{-rt}S_t(\lambda \sigma \text{d}t+\sigma \text{d}W_t).$$
We use Girsanov theorem to get a Q-BM: $\tilde{W_t}=W_t+\lambda t$
Now we replace it in the above equation, so we get $$\text{d}(e^{-rt}S_t) = \sigma e^{-rt}S_t\text{d}\tilde{W_t}.$$
How can I now prove this is a Q-mart?
I know I should use the Martingale Characterization, but I do not know, how to show that
$$E[\int_0^T(e^{-rt}S_t)^2]<\infty \quad \forall T>0,$$
which is the required condition for it.