Can I integrate an asymptotic expression?

Question

Suppose that $y(x; \epsilon)$ is a real-valued function of $x \in [a,b] \subset\mathbb{R}$ depending on a real parameter $\epsilon$, and that \begin{align} \int_a^b dx \ y(x; \epsilon) =& 1 && \text{for all } \epsilon. \end{align} If there is an asymptotic expression $g(x;\epsilon)$ to $y(x;\epsilon)$ such that \begin{align} y(x; \epsilon) \sim& g(x; \epsilon) &\text{ as } \epsilon \rightarrow& 0, \end{align} then does the following relation hold? \begin{align} \int_a^b dx\ g(x; \epsilon) \sim& 1 & \text{ as } \epsilon \rightarrow 0. \end{align}

---

Edit Wed Jul 13 21:58:07 CEST 2016

I erased all the addenda I made after posting the original question because these addenda are misleading.

Answer 1

I will use $t$ in this answer as opposed to $\epsilon$ because of limits. We have that $\lim_{t\to 0} \frac{g(x,t)}{y(x,t)}=1$, that is:

$$\forall \epsilon >0, \, \exists \delta>0,\, \forall t \text{ with } 0 < t < \delta,\\\left | \frac{g(x,t)}{y(x,t)}-1\right |\leq \epsilon \Longleftrightarrow (1-\epsilon)\cdot y(x,t)\leq g(x,t)\leq (1+\epsilon)\cdot y(x,t)$$

Can you see how this solves your problem?