Let $(\Omega, \mathcal{F}_{\infty}, \mathbb{P})$ be a probability space and let $(B_t)$ be a standard Brownian motion with its natural filtration $(\mathcal{F}_t)_{t\geq 0}$. Let $\mathbb{Q}$ be a probability measure which is absolutely continuous w.r.t $\mathbb{P}$ on $(\Omega, \mathcal{F}_{\infty})$ then Radon-Nikodym theorem yields the existence of measurable function $Z_{\infty} =\left. \frac{d\mathbb{Q}}{d\mathbb{P}}\right|_{\mathcal{F}_{\infty}}$ (the Radon-Nikodym derivative).
Now define $Z_t = \mathbb{E}[Z_{\infty}|\mathcal{F}_t]$ we want to show that $Z_t$ is a martingale with respect to $\mathcal{F}_t$. Let $s \leq t$ then
$$\mathbb{E}[Z_t | \mathcal{F}_{s}] = \mathbb{E}[\mathbb{E}[Z_{\infty}|\mathcal{F}_t] | \mathcal{F}_{s}]$$
the tower property of conditional expectation yields that
$$\mathbb{E}[\mathbb{E}[Z_{\infty}|\mathcal{F}_t] | \mathcal{F}_{s}] = \mathbb{E}[Z_{\infty}|\mathcal{F}_s]=Z_s.$$
Hence, $(Z_t)$ is a martingale.
Now we show that $Z_t = \left.\frac{d\mathbb{Q}}{d\mathbb{P}}\right|_{\mathcal{F}_t}$, indeed,
for any bounded $\mathcal{F}_t$-measurable function $f$ we have
\begin{align*}
\int_{\Omega}f \ d\mathbb{Q}=& \int_{\Omega} f Z_{\infty} \ d\mathbb{P} = \mathbb{E}_{\mathbb{P}}[fZ_{\infty}]\\=&\mathbb{E}[\mathbb{E}[fZ_{\infty} | \mathcal{F}_t]]= \mathbb{E}[f \mathbb{E}[Z_{\infty}|\mathcal{F}_t]] \\=& \mathbb{E}[fZ_t] = \int_{\Omega} fZ_t d\mathbb{P}.
\end{align*}
Hence, $Z_{\infty}d\mathbb{P} = Z_t d\mathbb{P}$ on $\mathcal{F}_t$ for each $t$.
