Can I integrate an asymptotic expression? 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.
 A: 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?
A: I am answering to my own question.
Proposition
First, let me restate the proposition with a little different symbols and slight generalization.
Suppose that $f(x,t)$ is a function on the domain $\{(x,t)|x\in [a,b], t\in D\}$ 
where $D$ is a neighbourhood of $t=t_0$. 
Suppose that $f(x,t)$ is asymptotically equivalent to another function $g(x,t)$ 
as $t\rightarrow t_0$ uniformly in $x$, which we write as
\begin{align}
f(x,t) \sim& g(x,t)&  \text{ as }& t\rightarrow t_0& 
\text{ uniformly for } x \in [a,b].
\end{align}
Suppose that the integral, 
\begin{equation}
I_g(t) := \int_a^b dx g(x,t),
\end{equation}
exists and is non-zero for all $t \in D$. (The integrability implies that $g(x,t)$ is 
bounded.)
Then the following asymptotic equivalence holds:
\begin{align}
\int_a^b dx f(x,t) \sim& \int_a^b dx g(x,t) &
\text{ as }& t\rightarrow t_0. 
\label{eq: asymp integ f g}
\end{align}
In fact, this proposition was stated in Sec. 6.2 of
C.M. Bender and S.A. Orszag, 
Advanced Mathematical Methods for Scientists and Engineers,
McGraw-Hill (1978)
and the proof was left as Prob. 6.3 therein.
In my original question I used $\epsilon$ instead of $t$ here as well as $g$ and $y$ instead of $f$ and $g$, respectively. I also restricted $I_g(t) \sim 1$ as $t\rightarrow t_0$.
Proof
First,
\begin{equation}\label{eq: proof step 1}
\begin{split}
\left|\int_a^b dx f(x,t) - \int_a^b dx g(x,t) \right|
=& \left|\int_a^b dx \left[f(x,t) - g(x,t)\right] \right| \\
\leq& \int_a^b dx \left|f(x,t) - g(x,t)\right| 
\end{split}
\end{equation}
From assumption of the uniform asymptotic equivalence of $f(x,t)$ and $g(x,t)$, for any positive number $\epsilon$, there exists another positive number $\delta$ independent of $x$ such that 
\begin{align}
|t-t_0|<&\delta& \Rightarrow&&
\left| f(x,t) - g(x,t)  \right| \leq& \epsilon |g(x,t)| &
\text{ for all } x \in& [a,b].
\label{eq: proof step 2}
\end{align}
Therefore, with the same $\epsilon$ and $\delta$, 
\begin{align}\label{eq: proof step 3}
|t-t_0| <& \delta& \Rightarrow&&
\int_a^b dx \left|f(x,t) - g(x,t)\right| 
\leq& \epsilon
\int_a^b dx \left|g(x,t)\right| .
\end{align}
Since we assumed that $I_g(t)$ exists, also $\int_a^b dx |g(x,t)|$ exists, i.e., finite. (See below for a proof.)
Let us write 
\begin{equation}\label{eq: proof step 4}
r:= 
\frac{ \int_a^b dx \left|g(x,t)\right|  }
     { |I_g(t)|  }.
\end{equation}
Since $I_g(t)$ is non-zero and finite, $0<r < \infty$.
(In fact, $1 \leq r < \infty$, but it does not matter. It only matters 
that $r$ is a positive real constant.)
We can say with the same $\epsilon$ and $\delta$ above that
\begin{align}
|t-t_0| <& \delta& \Rightarrow&&
\int_a^b dx \left|f(x,t) - g(x,t)\right| 
\leq& \epsilon r |I_g(t)|.
\end{align}
Combining the steps above together, for any positive number $\epsilon'$, we can take a positive number $\epsilon = \epsilon'/r$. Corresponding to this $\epsilon$, there exists $\delta$ such that
\begin{align}
|t-t_0| <& \delta& \Rightarrow&&
\left| \int_a^b dx f(x,t) - \int_a^b dx g(x,t)\right| 
\leq& \epsilon' \left|\int_a^b dx g(x,t)\right|.
\end{align}
By this we proved the proposition.
Riemann Integrability of a function and its modulus
Let $A:= [a,b] \subset \mathbb{R}$, and 
suppose $g(x): A \rightarrow \mathbb{R}$ is a bounded function.
If $g(x)$ is integrable over the domain $A$, then
$|g(x)|$ is also integrable over $A$.
This theorem was used in the proof of asymptotic equivalence above.
For the sake of completeness, here I write down a proof of this theorem.
This is merely a compact re-writing of relevant parts of 
J.K. Hunter, The Riemann Integral. 
First note that for every subinterval $I \subset A$, 
for every $x \in I$ and every $y \in I$, the following holds:
\begin{equation} 
\Big| |g(x)| - |g(y)| \Big| \leq |g(x) - g(y)|.
\end{equation} 
Note also that 
\begin{align}
|g(x) - g(y)| \leq& \sup_I g - \inf_I g, & 
\text{ for all } x,y\in& I
\end{align}
and that 
\begin{align}
\Big| |g(x)| - |g(y)| \Big| \geq& |g(x)| - |g(y)| \geq \sup_I|g| -\inf_I|g|, &
\text{ for all } x,y\in& I.
\end{align}
Therefore, 
\begin{equation}
\sup_I|g| -\inf_I|g| \leq \sup_I g - \inf_I g.
\tag{*}\label{*}
\end{equation}
Given a partition $P = \{I_k\}_{k=1,\dots,n}$ of the interval $A$, 
the upper and lower Riemann sums of $g$ on the partition $P$ are defined, 
respectively, as 
\begin{align}
U(g; P) :=& \sum_{k=1}^n |I_k| \sup_{I_k} g,  \\
L(g; P) :=& \sum_{k=1}^n |I_k| \inf_{I_k} g ,
\end{align}
where $|I_k|$ is the length of the subinterval $I_k$.
(If $I_k = [x_{k-1}, x_k]$, then $|I_k| = x_k -x_{k-1} >0$.)
Similarly, the upper and lower Riemann sums of $|g|$ on the partition $P$ are defined, 
respectively, as 
\begin{align}
U(|g|; P) :=& \sum_{k=1}^n |I_k| \sup_{I_k} |g|,  \\
L(|g|; P) :=& \sum_{k=1}^n |I_k| \inf_{I_k} |g|.
\end{align}
Note htat 
\begin{equation}
U(g; P) - L(g; P)
= \sum_{k=1}^n |I_k| \left[ \sup_{I_k} g  -\inf_{I_k} g \right],
\end{equation}
and that 
\begin{equation}
U(|g|; P) - L(|g|; P)
= \sum_{k=1}^n |I_k| \left[ \sup_{I_k} |g|  -\inf_{I_k} |g| \right].
\end{equation}
The necessary and sufficient condition that $g$ is Riemann integrable is that
for any $\epsilon>0$, there exists a partition $P$ such that 
\begin{equation}
U(g; P) - L(g; P) < \epsilon.
\end{equation}
(This is the Cauchy criterion of integrability. 
For a proof, see 
J.K. Hunter, The Riemann Integral.)
Then, with the same $\epsilon$ and $P$, 
using eq. (\ref{*}) we can say that 
\begin{equation}
\begin{split}
U(|g|; P) - L(|g|; P)
   =& \sum_{k=1}^n |I_k| \left[ \sup_{I_k} |g|  -\inf_{I_k} |g| \right]\\
\leq& \sum_{k=1}^n |I_k| \left[ \sup_{I_k} g  -\inf_{I_k} g \right]
= U(g; P) - L(g; P) 
< \epsilon.
\end{split}
\end{equation}
This implies that $|g|$ is integrable. 
Herewith we proved that if $g$ is Riemann integrable, then 
$|g|$ is also Riemann-integrable. 
