Consider an incomplete market $(\Omega,\mathcal F,\mathbb P)$ driven by a semimartingale $S=(S\_t)\_{t\in[0,T]}$. Under the [no free lunch under vanishing risk](http://en.wikipedia.org/wiki/No_free_lunch_with_vanishing_risk) (NFLVR) assumption, the set $\mathcal P^\ast$ of equivalent martingale measures under which $S$ is a sigma-martingale is non-empty:
$$
\mathcal P^\ast\neq\emptyset,\quad\mathbb P^\ast\sim\mathbb P,\enspace \forall \mathbb P^\ast \in \mathcal P^\ast.
$$


Denote densities 
$$
Z\_{\mathbb{P^\*}}=\frac{d\mathbb P^\ast}{d\mathbb P},\quad \text{for }\mathbb P^\ast\in \mathcal P^\ast,
$$


and the set of densities
$$Z\_{\mathcal{P^\ast}}=\{Z\_{\mathbb{P^\ast}}\ :\ \mathbb P^\ast\in \mathcal P^\ast\}$$


Now, a certain theorem (a generalized version of the Neyman-Pearson lemma for incomplete markets, see [Theorem 4.9 p.55 in this thesis](http://www.princeton.edu/~brudloff/Diss_Rudloff.pdf)) requires that $Z\_{\mathcal P^\ast}$ is compact in $L^1(\Omega,\mathcal F,\mathbb P)$. I would like to see how restricting this assumption is by trying to provide an incomplete market example (i.e., a model where $\mathcal P^\ast$ is not a singleton) where $Z\_{\mathcal P^\ast}$ is compact. However, I haven't been able to do so.


I first tried the discrete-time one-period trinomial model, in which the set of martingale measures and the set of associated densities represent convex polyhedrons. However, due to the measure equivalence requirement, the polyhedron is open (at the extreme points, one or more nodes will have zero probability, hence the extreme points are excluded along with the entire boundary). Therefore, it is not compact?


Next thing, I tried the continuous-time jump-diffusion model. The set of equivalent martingale measures can be parametrized with two real numbers, and the Radon-Nikodym derivative can be written down explicitly, however I have no idea how to approach the problem of verifying the compactness property.




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Could someone provide an example of an incomplete market (where the set of equivalent martingale measures is not a singleton) where the set of densities is indeed compact?




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Alternatively, can it be proved that $Z\_{\mathcal P^\ast}$ is compact if and only if $\mathcal P^\ast$ is a singleton?




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**Update**
As pointed below by @weakstar, if there exists a martingale measure $\mathbb Q$ that is absolutely continuous with respect to $\mathbb P$ but not equivalent to $\mathbb P$, a convex combination of $\mathbb Q$ and an equivalent martingale measure $\mathbb P^\ast\in\mathcal P^\ast$ would also be an equivalent martingale measure. In this case, it is possible to construct a sequence of densities $Z\_{\mathbb Q^\alpha}$ of equivalent martingale measures converging to density $Z\_{\mathbb Q}$ of a non-equivalent martingale measure which violates the closedness. This example naturally eliminates models like the multinomial model.



Compactness of the set of densities of equivalent martingale measures
The set $Z\_{\mathcal{P^\ast}}$ is *never* compact except in the case where it is a singleton (or empty). This is for the general case with $S=(S^1,S^2,\ldots,S^d)$ being an $\mathbb{R}^d$-valued semimartingale.


Let $\mathbb{Q}$ be an equivalent measure under which $S$ is a sigma-martingale, so $Z\_\mathbb{Q}\in\mathcal{P^\ast}$. If there existed any measure $\mathbb{Q}^\prime$ absolutely continuous, but not equivalent, to $\mathbb{P}$ with respect to which $S$ is a sigma-martingale then we can argue as in weakstar's answer. The measures $\mathbb{Q}\\_n=\frac1n\mathbb{Q}+(1-\frac1n)\mathbb{Q}^\prime$ are in $\mathcal{P}^\ast$, so $Z\_{\mathbb{Q}\\_n}=\frac1nZ\_\mathbb{Q}+(1-\frac1n)Z\_{\mathbb{Q}^\prime}$ are in $Z\_{\mathcal{P}^\ast}$. However, $Z\_{\mathbb{Q}^\prime}=\lim\_{n\to\infty}Z\_{\mathbb{Q}\\_n}$ in $L^1$. Hence, $Z\_{\mathcal{P^\ast}}$ is not closed in $L^1$ and cannot be compact. In general, you can show that if $\mathcal{P^\ast}$ contains more than one element, then there does exists an absolutely continuous measure which is not equivalent to $\mathbb{P}$ under which $S$ is a sigma-martingale, so $Z\_{\mathcal{P^\ast}}$ is not closed in $L^1$. The proof of this which I give here is rather long, because it is broken down into 4 separate cases and a tricky lemma. Maybe there is a more efficient proof, but I can't think of one now.


Suppose that $\mathbb{Q},\mathbb{Q}^\prime$ are distinct elements of $\mathcal{P^\ast}$. Replacing the original measure $\mathbb{P}$ by $\mathbb{Q}^\prime$ if necessary, without loss of generality we can suppose that $\mathbb{Q}^\prime=\mathbb{P}$. Furthermore, replacing $\mathbb{P}$ by $(\mathbb{P}+\mathbb{Q})/2$, we can suppose that $d\mathbb{Q}/d\mathbb{P}$ is uniformly bounded by 2. Also, replacing $S$ by $\int\varphi\,dS$ for some strictly positive predictable process $\varphi$, we can suppose that $S$ is an $\mathcal{H}^1$ martingale rather than just a sigma-martingale (see Proposition 2.6 of [The Fundamental Theorem Of Asset Pricing For Unbounded Stochastic Processes](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.16.9758)). That is, $\sup\_t\lVert S\_t\rVert$ is $\mathbb{P}$-integrable.


Define the $\mathbb{P}$-martingale
$$
U\_t=\frac{d\mathbb{Q}}{d\mathbb{P}}\Bigg\vert\\_{\mathcal{F}\_t}=\mathbb{E}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}\;\Bigg\vert\mathcal{F}\\_t\right].
$$
This is a uniformly bounded martingale with $\mathbb{E}[U\_t]=1$ such that $US$ is a martingale. We cannot have $U\_t=1$ almost surely for each $t$, as $\mathbb{P},\mathbb{Q}$ are assumed to be distinct.


Assume that the filtration is right-continuous and complete (you can always pass to the right-continuous version with no problems, and completeness is generally assumed for concepts such as stochastic integration to make sense). In that case, every martingale has a cadlag modification. So, take a cadlag modification for $U$. Then, $U$ is a strictly positive and non-constant martingale with $S,US$ both martingales. By integration by parts, this implies that the quadratic covariation $[U,S]$ is a local martingale. Suppose that we can find a nonnegative martingale $M$ which is not identically zero such that $MS$ is a martingale and $\mathbb{P}(M\_T=0) > 0$. Then, we can define an absolutely continuous but non-equivalent martingale measure $\mathbb{P}^\ast$ by
$$
\frac{d\mathbb{P}^\ast}{d\mathbb{P}}=M\_T/\mathbb{E}[M\_T].
$$
This would imply that $Z\_{\mathcal{P}^\ast}$ is not closed in $L^1$ by the argument above.
Let's consider the following four exhaustive (and increasingly exhausting) cases.


1. $U\_0$ is not almost-surely equal to 1. Then $\mathcal{F}\_0$ would contain a set $A$ with probability neither 0 or 1, so $M$ can be taken to be the (almost-surely constant) martingale
$$
M\_t=1\_A/\mathbb{P}(A).
$$
2. $U\_0=1$ and $\Delta U\_t\ge 0$ for all $t$. That is, $U$ only has positive jumps. As $U$ is assumed to be not almost-surely constant, there will exist $K\in(0,1)$ such that $\mathbb{P}(\inf\_tU\_t < K) > 0$. Let $\tau$ be the first time at which $U\_\tau\le K$, and $\tau=\infty$ if this never happens. As $U$ has nonnegative jumps, $U\_\tau=K$ whenever $\tau < \infty$. Then, the martingale $M$ can be defined by
$$
M\_t=U\_{t\wedge\tau}-K.
$$
Note that this hits zero whenever $\tau < \infty$, which happens with positive probability.
3. There exists a predictable stopping time $\tau > 0$ at which $\mathbb{P}(\Delta U\_\tau\not=0) > 0$. Then, as $U,S,US$ are martingales,
$$
\begin{align}
\mathbb{E}[\Delta U\_\tau \Delta S\_\tau\vert\mathcal{F}\\_{\tau-}]&=\mathbb{E}[\Delta(US)\\_\tau-U\_{\tau-}\Delta S\_\tau - S\_{\tau-}\Delta U\_\tau\vert\mathcal{F}\\_{\tau-}]\cr
&=0.
\end{align}
$$
As $U$ is bounded by $2$, $\Delta U\_\tau \ge -2$. Also, as $\Delta U\_\tau$ is not identically zero and $\mathbb{E}[\Delta U\_\tau\vert\mathcal{F}\\_{\tau-}]=0$, there is an $\epsilon > 0$ such that $\Delta U\_\tau\ge\epsilon$ with positive probability. Applying the lemma below to the $\mathbb{R}^{d+1}$-valued $\mathcal{F}\\_\tau$-measurable random variable $(1,\Delta S\_\tau)$ and nonnegative random variable $1+\Delta U\_\tau/2$, there exists a uniformly bounded $\mathcal{F}\\_{\tau-}$-measurable random variable $Y\ge1+\epsilon/2$, a nonnegative $\mathcal{F}\\_\tau$-measurable variable $X\le Y$ such that $\mathbb{P}(X=Y) > 0$ and,
$$
\begin{align}
&\mathbb{E}[X\Delta S\_\tau\vert\mathcal{F}\\_{\tau-}]=\mathbb{E}[(1+\Delta U\_\tau/2)\Delta S\_\tau\vert\mathcal{F}\\_{\tau-}]=0\cr
&\mathbb{E}[X\vert\mathcal{F}\\_{\tau-}]=\mathbb{E}[1+\Delta U\_\tau/2\vert\mathcal{F}\\_{\tau-}]=1.
\end{align}
$$
The martingale $M$ is then given by
$$
M\_t = 1 - 1\_{\{ t\ge\tau\}}(X-1)/(Y-1).
$$
Note that $M\_T=0$ whenever $X=Y$, which has positive probability, and that the quadratic covariation
$$
[M,S]\\_t=1\_{\{t\ge\tau\}}(X-1)\Delta S\_\tau/(Y-1)
$$
is a martingale so, by integration by parts, $MS$ is a martingale.
4. $U$ is quasi-left-continuous (i.e., $\Delta U\_\tau=0$ a.s. for each predictable stopping time $\tau$) and, with positive probability, $\Delta U\_t < 0$ for some $t\in(0,T]$. Then, there exists $\epsilon > 0$ so that, with positive probability, $\Delta U\_t < -\epsilon$ for some $t\in(0,T]$. Let $\tau$ be the first time at which $\Delta U\_\tau < -\epsilon$ and $\tau=\infty$ if this never happens. By the lemma below, there exists an $\mathcal{F}\\_{\tau-}$-measurable $Y\ge\epsilon$ and a nonnegative $\mathcal{F}\\_\tau$-measurable random variable $X\le Y$ such that $\mathbb{P}(X=Y) > 0$ and
$$
\mathbb{E}[X\Delta S\_\tau\vert\mathcal{F}\\_{\tau-}]=\mathbb{E}[-\Delta U\_\tau\Delta S\_\tau\vert\mathcal{F}\\_{\tau-}].
$$
Now, let $V$ be the step process $V\_t=1\_{\{t\ge\tau\}}(X+\Delta U\_\tau)$. The quadratic covariation
$$
[V,S]\\_t=1\_{\{t\ge\tau\}}(X+\Delta U\_\tau)\Delta S\_\tau
$$
is a martingale. As $\mathbb{P}(\sigma=\tau < \infty)=0$ for each predictable $\sigma$, $\tau$ is *totally inadmissible*. Therefore, it has a compensator $V^p$, which is a continuous finite variation adapted process starting from zero such that $V-V^p$ is a martingale. Consider the martingale
$$
N\_t=U\_{t\wedge\tau}+V^p\_t-V\_t.
$$
Then, $[N,S]=[U,S]^\tau-[V,S]$ is a local martingale. Also, $\Delta N\_t\ge-\epsilon$ for $t < \tau$ and $\Delta N\_\tau=-X\ge-Y$ (and equality holds with positive probability). As $Y$ is $\mathcal{F}\\_{\tau-}$-measurable, there is a predictable process $\xi$ with $\xi\_\tau=Y$. Replacing $\xi$ by $\xi\vee\epsilon$, we can suppose that $\xi\ge\epsilon$. Consider the martingale
$$
\tilde N=\int\xi^{-1}\,dN.
$$
Then $\Delta\tilde N\_t=\xi\_t^{-1}\Delta N\_t\ge-1$ and $\Delta\tilde N\_\tau=-1$ with positive probability. Let $M$ be the solution to the SDE
$$
M\_t = 1+\int\_0^tM\_{s-}\,d\tilde N\_s.
$$
This is the *Doléans exponential* of $\tilde N$ and is given explicitly by
$$
M\_t=\exp\left(\tilde N\_t-\frac12[\tilde N]\\_t\right)\prod\_{s\le t}e^{-\Delta\tilde N\_s+\frac12(\Delta\tilde N\_s)^2}\left(1+\Delta\tilde N\_s\right).
$$
As $\Delta\tilde N\ge-1$, this is nonnegative. Also, $M\_T=M\_\tau=0$ whenever $\Delta\tilde N\_\tau=-1$, which occurs with positive probability. As $dM\_t=M\_{t-}\,d\tilde N\_t$ and $d[M,S]\\_t=M\_{t-}\,d[\tilde N,S]\\_t$, $M$ and $[M,S]$ are local martingales so, by integration by parts, $MS$ is a local martingale. By localization (replacing $M$ by $M^\sigma$ for a suitable stopping time $\sigma$ with $\mathbb{P}(\sigma\ge\tau) > 0$), we can assume that $M$ and $MS$ are both proper martingales. So, $M$ satisfies the required properties.


That's it! There does exist an absolutely continuous but not equivalent martingale measure. It just remains to state and prove the following lemma which was used in cases 3 and 4.



> 
> **Lemma:** Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G}$ be a sub-σ-algebra of $\mathcal{F}$, $S$ be an integrable $\mathbb{R}^d$-valued random variable and $Z$ be a uniformly bounded nonnegative random variable with $\mathbb{P}(Z\ge K) > 0$ for some $K > 0$. Then, there exists a nonnegative random variable $X$ and a uniformly bounded $\mathcal{G}$-measurable random variable $Y\ge K$ such that $X\le Y$, $\mathbb{P}(X=Y) > 0$ and
> $$
> \mathbb{E}[XS\vert\mathcal{G}]=\mathbb{E}[ZS\vert\mathcal{G}].
> $$
> 
> 
> 




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To prove the lemma, I'll make use of the relatively simple fact that if $\mu$ is a probability measure on $\mathbb{R}^d$ and $S\subseteq\mathbb{R}^d$ is Borel with $\mu(S)=1$, then $\int x\,d\mu(x)$ lies in the convex hull of $S$ (note: this only works in finite dimensions. In general Banach spaces you are only guaranteed that it lies in the closure of the convex hull). Let $V$ be an affine subspace with minimal dimension such that $\mu(V)=1$. Then, $y\equiv\int x\,d\mu(x)\in V$. Suppose that it did not lie in ${\rm conv}(S)$. Then, by the Hahn-Banach theorem/separating hyperplane theorem, there exists a nontrivial affine map $L\colon V\to\mathbb{R}$ such that $L(x)\ge0$ for $x\in S$ and $L(y)=0$. However, $L(y)=\int\\_V L(x)\,d\mu(x)$. For this to be zero, $L(x)$ must be zero $\mu$-almost everywhere, so $\mu(W)=0$ where $W=L^{-1}(0)$ contradicting the minimality of $V$. So $y\in{\rm conv}(S)$.


Next, if $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space and $U\colon\Omega\times(0,1)\to\mathbb{R}^d$ is jointly measurable such that $U(\cdot,x)$ is $\mathbb{P}$-integrable for each $x$, then we can define a random variable $V(\omega)=\int\_0^1U(\omega,x)\,dx$. By the paragraph above, $V(\omega)$ lies in the convex hull of $\{U(\omega,x)\colon x\in(0,1)\}$ for each $\omega$. The convex combinations can be done in a measurable fashion; there are sequences $p\_n,x\_n$ of random variables such that $p\_n$ are nonnegative, eventually zero with $\sum\_np\_n=1$, $x\_n\in(0,1)$ and
$$
V(\omega)=\sum\_np\_n(\omega)U(\omega,x\_n(\omega))
$$
for $\mathbb{P}$-almost every $\omega$. That this can be done in a measurable way is a consequence of the [measurable section theorem (broken link)](https://planetmath.org/MeasurableSectionTheorem) (that's the planetmath link I wrote, see also Donald L. Cohn, Measure theory. Birkhäuser, 1980).


Now, moving onto the proof of the lemma. Let $Y$ be the essential infimum of all $\mathcal{G}$-random variables bounded below by $Z\vee K$. By definition, $K\le Y$ and $Z\le Y$. If $\mathbb{P}(Z=Y) > 0$ then we can set $X=Z$ and we are done. Otherwise, for each $x\in(0,1)$, define the random variable
$$
U\_x=\mathbb{E}\left[1\_{\{Z\ge xY\}}YS\vert\mathcal{G}\right].
$$
This can be chosen to be jointly measurable (e.g., take it to be finite variation in $x$) and,
$$
\int\_0^1U\_x\,dx=\mathbb{E}\left[\int\_0^11\_{\{Z\ge xY\}}YS\,dx\big\vert\mathcal{G}\right]=\mathbb{E}[ZS\vert\mathcal{G}].
$$
By the discussion above, this can also be written in the form $\sum\_np\_nU\_{x\_n}$ for $\mathcal{G}$-measurable random variables $p\_n\ge0$, $x\_n\in(0,1)$ where $\sum\_np\_n=1$ and $p\_n$ is eventually zero. Set
$$
X=\sum\_np\_n1\_{\{Z\ge x\_nY\}}Y.
$$
This satisfies the required identity. Letting $x\_\ast(\omega)$ be the maximum of $\{x\_n(\omega)\colon p\_n(\omega) > 0\}$, then $(x\_\ast Y)\vee K < Y$ with positive probability so, by the definition of $Y$ as an essential infimum, $Z > x\_\ast Y$ with positive probability. However, in this case, $X=Y$.


