I am a physicist caught in the following situation:
I have two probability measures $\mathbb{P}\_1$ and $\mathbb{P}\_2$ and have to deal with the following integral where $X\_i$ are random iid:


$$\int\_{B} \mathbb{P}\_1\left(\frac{1}{N} \sum\_{i=1}^N X\_i \ge x\right) d\mathbb{P}\_2(x).$$


I was able to obtain a large deviation principle for $X$, i.e. I was able to show that 


$$\frac{1}{N} \log\left( \mathbb{P}\_1\left(\frac{1}{N} \sum\_{i=1}^N X\_i \ge x\right)\right) = -I(x).$$


Now it is tempting to replace $\mathbb{P}\_1\left(\frac{1}{N} \sum\_{i=1}^N X\_i \ge x\right) $ in the above integral by $e^{-N I(x)}$ and compute 
$$\int\_{B} e^{-NI(x)} d\mathbb{P}\_2(x).$$


However, I do not know in which sense this now close to the expression I am looking for? 


Are there any standard bounds to estimate the error between my approximation and the object I am interested in?



Large deviations for integrands
It all depends on the shape of $P\_2$ and on the assumptions you put on $X\_i$. In what follows I'll assume that $\Lambda(\lambda)=\log E\_1 e^{\lambda X\_1}$ is finite 
for all $\lambda$. I will also assume that $E\_1X\_i=0$. Further I will assume that 
$P\_2$ is supported on $R$ with density $f$. 


Case 1: $B\cap (-\infty,0]>0$. In that case your formula may be false, and the answer is essentially $P\_2(B\cap (-\infty,0))$. 


Case 2: (which is what I guess you had in mind): $B\cap (-\infty,a)=\emptyset$ for some $a>0$. In that case, use Bahadur-Rao to approximate, for $x\in B$,
$P\_1(N^{-1}\sum\_{i=1}^N X\_i>x)\sim C(x)e^{-NI(x)}/\sqrt{N}$, with explicit $C=C(x)$. So the expression you write is correct at the exponential scale but if you meant up to precise asymptotics, you are missing a constant multiple and 
a factor $\sqrt{N}$. The constant $C(x)$ depends on whether the law of $X\_1$ is lattice or not. See the original paper of Bahadur-Rao or Theorem 3.7.4 in Dembo-Zeitouni's large deviations book.


One can also deal with other cases ($B$ touching $0$ with density of $P\_2$ vanishing there, etc.) but I'll stop here.


