I need some help understanding a note given in a lot of papers I've read.
Let $(\Omega,\mathcal{F},P)$ be a complete probability Space, $\mathbb{F} = (\mathcal{F}_t)_{t\in[0,T]}$ a given filtration with usual conditions, $S$ be a locally bounded semi-martingale and $$M_a = \{ Q \ll P \; | \; S \text{ is a locale } (Q,\mathbb{F})-\text{martingale}\}$$ the set of all absolutely continuous martingale measures.
Now I found a lot of papers claiming it's natural to assume that the set $$M_e = \{ Q \in M_a \; | \; H[Q|P] < +\infty \}$$
is non empty where
$$H[Q|P] = \begin{cases} E_P\Big[\frac{dQ}{dP}\log\frac{dQ}{dP}\Big] & \mbox{ if }Q \ll P \\ +\infty & \mbox{ otherwise} \end{cases} $$
is the relative entropy of $Q$ w.r.t $P$.
But I see no reason why $M_e \not= \emptyset$ should hold.
Maybe someone has a hint or link for me.
edit 09.12.2019: To point out my main issue here (see comments to second answer): It's totally clear to me why the set $M_a$ of absolutely continuous (locale) martingale measures is not empty if we have an arbitrage free (in the NFLVR sense) market. But why does at least one of these measures have a finite entropy related to our initial measure $P$?
Greetings
$M_e$ contains $P$, because $P\in M_a$ and $\frac{dP}{dP}=1$ and therefore $H[P|P]=0.$
I will show that $M_e$ contains more than just $P$ in the case where $S$ is a Brownian motion. In that case, Girsanov's theorem tells us that the elements in $M_a$ are the probability measures $Q$ under which $S$ is a Brownian motion with an added continuous drift $b$, and that $\frac{dQ}{dP}$ satisfies $\frac{dQ}{dP}=D_T$, where \begin{equation*} D_t = \exp\bigg( \int^t_0b(u)dS_u - \tfrac{1}{2}\int^t_0 b(u)^2 du \bigg) \end{equation*} for any $t\in[0,T]$, under the condition that $D$ is a martingale. One sufficient condition that $D$ is a martingale is the Novikov condition, which requires that \begin{equation*} E_P\bigg[\exp\bigg(\tfrac{1}{2}\int^T_0 b(u)^2 du\bigg)\bigg]<\infty. \end{equation*} If $D$ is a martingale, then substitution of $\frac{dQ}{dP}$ in the relative entropy yields \begin{equation*} \begin{split} E_P\bigg[\frac{dQ}{dP}\log\frac{dQ}{dP}\bigg] = E_Q\bigg[\log\frac{dQ}{dP}\bigg] &= E_Q\bigg[\int^T_0 b(u)dS_u\bigg] - E_Q\bigg[\tfrac{1}{2}\int^T_0 b(u)^2 du\bigg] \\ &= E_P\bigg[\int^T_0 b(u)dS_u\bigg] + E_Q\bigg[\tfrac{1}{2}\int^T_0 b(u)^2 du\bigg] \end{split} \end{equation*} (note in the last equality the one change of measure from $Q$ to $P$) where it is used that $S$ is a Brownian motion plus drift $b$ on $(\Omega,\mathcal{F},Q)$. If $b$ satisfies \begin{equation*} E_Q\bigg[\int^T_0 b(u)^2 du\bigg]<\infty, \end{equation*} then the stochastic integral $\int^T_0 b(u)dS_u$ will have zero expectation, and it follows that $H[Q|P]$ is finite.
You might find the following article interesting: http://jbierkens.nl/pubs/2014-scl.pdf It considers the above in more detail and in the context of control theory, where the drift $b$ acts as a control and $H[Q|P]$ as a control cost.