The answer to this question should be fairly easy, but I can not just see it.
I want to say something like: let us consider a measure $P_{\theta}\in\mathcal{P}$ where $\mathcal{P}=\left\{P_{\theta}:\frac{dP_{\theta}}{d\lambda}=I_{(\theta-0.5,\theta+0.5)}(x) \right\}$, that is, the family of uniform distributions in $(\theta-0.5,\theta+0.5)$, where $\frac{dP_{\theta}}{d\lambda}$ is the Radon-Nikodym derivative of $P_{\theta}$ w.r.t. the Lebesgue measure.
That is, I am assuming $\mathcal{P}$ is dominated by the Lebesgue measure and the Radon- Nikodym derivative of $P_{\theta}$ w.r.t. the Lebesgue measure $\lambda$ is \begin{equation} \frac{dP_{\theta}}{d\lambda}=I_{(\theta-0.5,\theta+0.5)}(x) \end{equation}
However, if I want to prove that $\mathcal{P}$ is not an exponential family, one way to prove it is to show that there does not exist a dominating measure that results in such a Radon-Nikodym derivative.
The argument is as follows. If there exists such a measure, let us say $\nu$, with Radon-Nykodim derivative \begin{equation} \frac{dP_{\theta}}{d\nu}=I_{(\theta-0.5,\theta+0.5)}(x) \end{equation}
then for each $t\geq 0$ there is a $\theta<t-0.5$ such that $P_\theta[[t,\infty)]=0$, which, along with the existence of the previous Radon-Nikodym derivative implies $\nu[[t,\infty)]=0$, which implies that $\nu[[0,\infty)]=0$. Following a symmetric argument, $\nu[(-\infty,0]]=0$ and then $\nu=0$.
Therefore, there is no such a measure $\nu$ that gives us the Radon-Nykodim derivative, which is requirement for the exponential family.
I am sure I am missing something.
Any hints?
The question in the post is not the question in the title. The answer to the question in the title is "yes, obviously" since for every $\theta$, $P_\theta\ll\lambda$ by definition. The answer to the question in the post is "no, $\mathcal P$ is not an exponential family". To see this, recall that the support of every measure in an exponential family does not depend on the measure, here $[h\ne0]$ should be independent of $\theta$.
But $P_0((1/2,1))=0$ while $P_1((1/2,1))\ne0$ hence $\mathcal P$ is not an exponential family. In fact, for every $t\ne s$, $\{P_t,P_s\}$ is not an exponential family.