A stochastically increasing exponential family for which $\lim_{\theta\rightarrow\inf\Theta}\mbox{P}_\theta(X\leq x)\neq 1$ Question
A little something that I've been wondering about for a while:
Let $P_\theta$ be a stochastically increasing (one-parameter) exponential family on the sample space $\mathcal{X}$ with $\Theta\subset\mathbb{R}$ being its natural parameter space, i.e. $\Theta$ being the set of values for which the cdf $F_\theta$ defines a probability measure. Is it always true that
$$F_\theta(x)\nearrow 1\qquad\mbox{as}\qquad\theta\searrow \inf\Theta,$$
and
$$F_\theta(x)\searrow 0\qquad\mbox{as}\qquad\theta\nearrow \sup\Theta$$
for all $x\in\mathcal{X}$ which are not in the boundary of $\mathcal{X}$?

Definitions, an example, speculation
A distribution $P_\theta$ over $\mathcal{X}$ parametrized by $\theta\in\Theta$ is stochastically increasing if, for fixed but arbitrary $x\in\mathcal{X}$, $F_\theta(x)$ is decreasing in $\theta$, where $F_\theta(x)=\mbox{P}_\theta(X\leq x)$.
An example is the binomial distribution, where $\mathcal{X}=\{0,1,2,\ldots,n\}$ and the natural parameter space is $\Theta=(0,1)$.
$$F_\theta(x)=\sum_{k\leq x}{\binom{n}{k}}\theta^k(1-\theta)^{n-k}$$
is decreasing in $\theta$. 
Example:

In this setting we have
$$F_\theta(x)\nearrow 1\qquad\mbox{as}\qquad\theta\searrow \inf\Theta=0,$$
and
$$F_\theta(x)\searrow 0\qquad\mbox{as}\qquad\theta\nearrow \sup\Theta=1$$
for all $x$ not on the boundary of $\mathcal{X}$. At the boundary, i.e. when $x\in \{0,n\}$, only one of the limits is attained.
Note that if we restrict the parameter space to be a proper subset of $(0,1)$, such as $(0.2,0.8)$, this is no longer true: $F_\theta(5)\nearrow F_{0.2}(5)=0.174\ldots<1$ as $\theta\searrow \inf\Theta=0.2$.
This property seems to hold for all commonly used exponential families, so my guess would be that if a counterexample exists it has to be somewhat pathological. It could perhaps involve functions that cause $\exp(<\theta,T(x)>)$ to blow up for some finite $(\theta,x)$, making its integral infinite. In some sense this would make the parameter space "restricted".
A "trivial" example is perhaps the Bernoulli(p) distribution, since its sample space equals its own boundary, so that only one of the limits can be attained for each point in the sample space. But that's a wee bit boring, and I'd much rather have example where at least one point of the sample space is not in the boundary.
 A: Based on comments, we are not restricted to considering the natural parameter but are allowed to use the general form
\begin{equation}
f_\theta(x) \propto h(x) e^{T(x)\eta(\theta)}
\end{equation}
In this case, it is possible to 'cheat' by constructing $\eta$ so that some values in the natural space of $\eta$ are not reached with any $\theta$. For example, let
\begin{equation}
f_\theta(x) \propto e^{-x\,\frac{1}{1+e^\theta}},~x\geq0
\end{equation}
This is a (strangely parametrized) exponential distribution with CDF 
\begin{equation}
F_\theta(x) = 1 - e^{-x / ({1+e^\theta})},
\end{equation}
which is a decreasing function for $\theta$ as desired. However, 
\begin{equation}
\lim_{\theta\rightarrow -\infty} F_\theta(x) = 1-e^{-x} < 1.
\end{equation}
The case with natural parametrization $(\eta(\theta) = \theta)$ remains open (at least to me). 
A: If discontinuities in the density is allowed, it is possible to construct a distribution that repeats itself over two consecutive intervals, which bounds the CDF at the first interval to $[0,0.5]$ and at the second interval to $[0.5,1]$. For example, let 
\begin{equation}
h(x) = 1, x\in [0,4]
\end{equation}
\begin{equation}
T(x) = \left\{
        \begin{array}{ll}
            1 & \quad x \in [0,1) \cup[2,3) \\
            2 & \quad x \in [1,2) \cup [3,4]
        \end{array}
    \right.
\end{equation}
Now, for a density proportional to $h(x)e^{\theta T(x)}$, the CDF is 
\begin{equation}
F_\theta(x) = \left\{
        \begin{array}{ll}
            \frac{1}{2}\,\frac{x}{1+e^\theta} & \quad x \in [0,1) \\
            \frac{1}{2}\, \frac{1 + (x-1)e^\theta}{1+e^\theta} & \quad x \in [1,2) \\
            \frac{1}{2} + \frac{1}{2}\,\frac{x-2}{1+e^\theta} & \quad x \in [2,3) \\
            \frac{1}{2} + \frac{1}{2}\, \frac{1 + (x-3)e^\theta}{1+e^\theta} & \quad x \in [3,4) \\
        \end{array}
    \right.
\end{equation}
which is decreasing as a function of $\theta$ for any $x$ in the sample space, but, for example,
\begin{equation}
\lim_{\theta \rightarrow \infty} F_\theta(2.1) = \frac{1}{2}.
\end{equation}
