The Radon-Nikodym derivative $\frac{dQ_T}{dQ_B}$ to change the measure from $Q_B$ to $Q_T$ can be obtained by considering the expectations
$\mathbb{E}^{Q_T} [\frac{P(t; T)V(T)}{P(T; T)}|\mathcal{F}_t]= \mathbb{E}^{Q_B} [\frac{P (t; T)V (T)}{P(T; T)}\frac{dQ_T}{dQ_B} |\mathcal{F}_t]=\mathbb{E}^{Q_B}[\frac{B(t)V(T)}{B(T)}|\mathcal{F}_t]$
where the first equality comes from applying the Radon-Nikodym derivative so that $E^{Q_T} [X] = E^{Q_B}[X \frac{dQ_T}{dQ_B}]$, and the second equality comes from Equation (11).
Hence we have $\frac{dQ_T}{dQ_B} =\frac{B(t)/B(T)}{P(t; T)/P(T; T)}$.
I do not understand why the bold part hold. It is concluded from $\mathbb{E}^{Q_B} [\frac{P (t; T)V (T)}{P(T; T)}\frac{dQ_T}{dQ_B} |\mathcal{F}_t]=\mathbb{E}^{Q_B}[\frac{B(t)V(T)}{B(T)}|\mathcal{F}_t]$ that $\frac{dQ_T}{dQ_B} =\frac{B(t)/B(T)}{P(t; T)/P(T; T)}$.
Here $(\mathcal{F}_t)_{t\geq 0}$ is a filtration.
We have that two $\mathcal{F}_t$ random variables which have the same conditional expectation for every $t<T$. Why does this imply that the two random variables are equal (a.s.)?