I approximate a probability distribution $P\_x(x)$ with a $P\_x^{app}(x)$,  

such that $P\_x(x)-P\_x^{app}(x) = O(\epsilon)$ uniformly in x, where epsilon is a small positive quantity.  

Consider the generic observable $Y = Y(x)$; $Y$ is a generic function of the random variable $x$, and so it is also a random variable.
If I want $E(y) \equiv \int P\_x(x') Y(x') dx' $ I expect an expression as:  

$E(y) = E^{app}(y) + O(\epsilon)$, where $E^{app}(y) \equiv \int P\_x^{app}(x') Y(x') dx' $.  

How can one shows this from a formal point of view ?



Perturbative approach starting from a probability distribution approximated form
$\newcommand\ep\epsilon\newcommand\R{\mathbb R}$Translated into a standard language of probability theory, your question appears to be as follows:



> 
> Let $p$ and $q$ be two probability density functions (say on $\R$) such that
> $$|p-q|\le\ep\tag{0}$$ for some real $\ep>0$. Let $Y$ be a Lebesgue-measurable function from $\R$ to $\R$. Does it then follow that
> $$\Big|\int pY-\int qY\Big|\le C\ep\tag{1}$$ for some real $C\ge0$?
> 
> 
> 


The answer is this: inequality (1) holds for some real $C>0$ and
all $p$ and $q$ satisfying condition (0) if and only if
$$C\_Y:=\int|Y|<\infty;$$
moreover, the best possible constant factor $C$ in (1) is $C\_Y$.


Indeed, on the one hand, if (0) holds, then
$$\Big|\int pY-\int qY\Big|=\Big|\int(p-q)Y\Big|\le\int|p-q|\,|Y|\le\int\ep\,|Y|=C\_Y\ep,$$
so that (1) holds with $C=C\_Y$.


On the other hand, suppose that
$$p=\ep\,1\_{[0,1/\ep]},\quad q=\ep\,1\_{[-1/\ep,0]},$$
$Y\ge0$ on the interval $[0,1/\ep]$ and $Y=0$ outside this interval. Then (0) holds, and
$$\Big|\int pY-\int qY\Big|=\int pY=\ep\int Y=C\_Y\ep.$$


So, $C\_Y$ is the best possible constant factor $C$ in inequality (1) given condition (0).


