Proving the MVUE is the following I am stuck on the following question and I was wondering if can get some help. 
Let $f(x;\theta) = g(\theta)h(x),\ a(\theta) \leqslant x \leqslant b(\theta)$ with $a(\theta)$ decreases and $b(\theta)$ increases with $\theta$ and $g(\theta)$ differentiable with random sample $X_1,...,X_n$ and $X_{(1)}<...<X_{(n)}$ the order statistics. Consider the statistic $T = max(a^{-1}(X_{(1)}), b^{-1}(X_{(n)}))$
I have to prove that the MVUE of $\theta$ is $\hat{\theta}=T-\frac{g(T)}{ng'(T)}$
I have done the completeness and sufficiency tests for T and since $\hat{\theta}$ is a one-to-one function of T, then it is also sufficient and complete by known theorems. I am now trying to calculate $E(T)$ using the following answer to post here: Find the unique MVUE
However, for this question, they were able to use the properties of the uniform to find the CDF but for my particular problem I do not believe that I should use that arguement. I was wondering if I can get some hints as to come up with the suitable CDF?
EDIT
I am up to this part:
\begin{equation}
F(t) = P(\hat{\theta} \leqslant t)
= P(a(t)\leqslant x_{(1)} \leqslant x_{(2)} \leqslant...\leqslant x_{(n)} \leqslant b(t))
\end{equation}
But again don not know how to proceed
EDIT 
so I believe to get the result I have to do integration by parts to get $E(T)=\theta+E\left(\frac{g(T)}{n g^{\prime}(T)}\right)$
Since $$
P_{\theta}(T \leq t)=(g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n}
$$
and $$
f_{T}(t)=n(g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n-1}\left[h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t)\right] \mathbf{1}_{0<t<\theta}
$$
I have the following results
$$
\begin{array}{c}{E(T)=\int_{a(\theta)}^{b(\theta)}\left(\operatorname{tn}(g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n-1}\left[h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t)\right] \mathbf{1}_{0<t<\theta}\right) d t} \\ {u=\operatorname{tn}(g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n-1}} \\ {\quad d u=n g(\theta)^{n}\left[\int_{a(t)}^{b(t)} h(x) d x+s\left(\left(h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t)\right)\right] d t\right.} \\ {v=\int_{a(\theta)}^{b(\theta)} h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t) d t}\end{array}
$$
but
$$
v=\int_{a(\theta)}^{b(\theta)} h(b(t)) b^{\prime}(t) d t-\int_{a(\theta)}^{b(\theta} h(a(t)) a^{\prime}(t) d t
$$
but doing the substitution $r=b(t), d r=b^{\prime}(t) d t$ and $s=a(t), d s=a^{\prime}(t) d t$ and
$\int_{a(\theta)}^{b(\theta)} h(x) d x=\frac{1}{g(\theta)},$ v should be 0 so i am stumped. I was hoping for some hint as to create $u v=\theta$
 A: Possible way to proceed:
For any $0<t<\theta$, distribution function of $T$ is
\begin{align}
P_{\theta}(T\le t)&=P_{\theta}\left[a(t)\le X_{(1)},X_{(n)}\le b(t)\right]
\\&=P_{\theta}\left[a(t)\le X_1,X_2,\ldots,X_n\le b(t)\right]
\\&=\left[P_{\theta}(a(t)\le X_1\le b(t))\right]^n
\\&=(g(\theta))^n \left(\int_{a(t)}^{b(t)}h(x)\,dx\right)^n
\end{align}
Density of $T$ is therefore $$f_T(t)=n(g(\theta))^n \left(\int_{a(t)}^{b(t)}h(x)\,dx\right)^{n-1}\left[h(b(t))b'(t)-h(a(t))a'(t)\right]\mathbf1_{0<t<\theta}$$
Since you say you have verified $T$ is complete sufficient, any function of $T$ that is unbiased for $\theta$ would be the UMVUE of $\theta$.
So you might set up $E_{\theta}\left[ k(T)\right]=\theta\ldots (*)$ for some function $k$ of $T$ and solve for $k$. One way to do that is to differentiate both sides of $(*)$ with respect to $\theta$. I would leave the details to you.
Keep in mind that $$\int_{a(\theta)}^{b(\theta)} f(x;\theta)\,dx=1\implies \int_{a(\theta)}^{b(\theta)}  h(x)\,dx=\frac{1}{g(\theta)}$$
I don't see how to readily verify whether the proposed estimator $T-\frac{g(T)}{ng'(T)}$ is unbiased for $\theta$.
