Let $\lambda$ be a probability measure on $\Omega$ (finite), with filtration $\{\mathcal{F}_t\}$. Define $\nu(X) = \lambda\left(X\frac{d\nu}{d\lambda}\right)$, where $\frac{d\nu}{d\lambda}$ is a random variable i.e., $\nu(\omega) = \lambda(\omega)\frac{d\nu}{d\lambda}(\omega)$, all $\omega\in\Omega$. Show that $$E\nu[X|\mathcal{F_t}] = \frac{E_{\lambda}\left[X\frac{d\nu}{d\lambda}|\mathcal{F_t}\right]}{E_{\lambda}\left[\frac{d\nu}{d\lambda}|\mathcal{F}_t\right]}$$
Recall from the second fundamental theorem of asset pricing $$\frac{d\nu}{d\lambda} = \frac{S_T^{0}}{\lambda(S_T^{0})}$$ if $S_T^{0}$ is a constant then $$\frac{d\nu}{d\lambda} = 1 \ \ \Rightarrow \ \ \lambda = \nu$$ The change of measure formula is $$E_{\nu}[X] = E_{\lambda}\left[X\frac{d\nu}{d\mu}\right]$$
For some attainable claim $X$ let $\phi$ be a self financing strategy replicating $X$ then by the first fundamental theorem of asset pricing $$V_t(\phi) = E_{\nu}\left[X\frac{S_t^{0}}{S_T^{0}} |\mathcal{F_t}\right]$$
I am pretty sure the result will follow from one of these fundamental theorems of asset pricing but I am not sure where to go from here. Sorry for the messy start, also if you need me to write the three fundamental theorems I would be happy to do so. Any comments or suggestions is greatly appreciated.
Alternative Solution - For all $\omega\in \Omega$, let $\mathcal{F}_t(\omega) = \mathcal{F}_t$ be the partition element containing $\omega$. Then
\begin{align*} E_{\nu}[X|\mathcal{F}_t](\omega) &= \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\nu(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \nu(\omega)}\\ &= \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)}\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}\\ &= \frac{\left( \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)} \right )}{\left(\frac{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)} \right )}\\ &= \frac{E_{\lambda}\left[X\frac{d\nu}{d\lambda}|\mathcal{F}_t\right]}{E_{\lambda}\left[\frac{d\nu}{d\lambda}|\mathcal{F}_t\right]} \end{align*}
My apologies there is something wrong with the latex generator on here, I copied and pasted the above in a few online latex generators and it works fine, do not know what the issue is here...\begin{align*} E_{\nu}[X|\mathcal{F}_t](\omega) &= \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\nu(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \nu(\omega)}\\ &= \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)}\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}\\ &= \frac{\left( \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)} \right )}{\left(\frac{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)} \right )}\\ &= \frac{E_{\lambda}\left[X\frac{d\nu}{d\lambda}|\mathcal{F}_t\right](\omega)}{E_{\lambda}\left[\frac{d\nu}{d\lambda}|\mathcal{F}_t\right](\omega)} \end{align*}
