This is not a homework or assignment exercise.
I'm trying to evaluate $\displaystyle \ \ I := E_\beta \big[\frac{1}{\beta(T_0)} K \mathbf{1}_{\{B(T_0,T_1) > K\}}\big]$, where $\beta$ is the savings account, $B$ is the present value (at some time $t$) of the zero coupon bond paying $\$1$ at maturity, $\mathbf{1}$ is the indicator function, $r(s)$ is the short rate, and $f(t,T)$ is the instantaneous forward rate (i.e. $f(t,T,T)$). Expanding on this a bit more:
$$ \displaystyle \ \ \beta(T_0) = e^{\int_0^{T_0} r(s)d(s)}$$
$$ \displaystyle \ \ B(t,T) = e^{-\int_t^T f(t,s)d(s)}$$
The first step is to do a measure change as follows:
$$\displaystyle \ \ I = E_{T_0} \big[\frac{1}{\beta(T_0)} K \mathbf{1}_{\{B(T_0,T_1) > K\}} \frac{\beta(T_0)B(0,T_0)}{\beta(0)B(T_0,T_0)}\big]$$
My question is why is this correct? For a density process to be a valid change of measure we need to to be a Doleans-Dade exponential local martingale, but I can't see why this is the case in this example, because:
$$\displaystyle \ \ \frac{\beta(0)B(T_0,T_0)}{\beta(T_0)B(0,T_0)} = 1$$
which is not in the usual form of a Doleans exponential?
