I am reading a proof in an introduction course to homogenization of partial differential equations. We want to prove the following statement:
Let $f\in L^{\infty}(\mathbb{R})$ be a $1$-periodic function. Define $f_{\epsilon}(x):=f\left(\tfrac{x}{\epsilon}\right)$. For any open interval $I\subset R$, $$\lim_{\epsilon\to 0}f_{\epsilon}=\langle f\rangle:=\int_{0}^{1}f(x)\text{d}x$$ in the sense of the weak$^{\ast}$ convergence.
First problem: where do we use $I$?
Proof. By definition of the weak$^{\ast}$ convergence, we have to prove that for any $g\in L^{1}(I)$, $$\lim_{\epsilon\to 0}\int_{I}f_{\epsilon}(x)g(x)\text{d}x=\langle f\rangle \int_{I}g(x)\text{d}x$$
The proof then use the density of simple functions in $L^{1}(I)$: for any $k\in\mathbb{N}_{0}$, there exists a piecewise constant function $g_{k}$ such that $$\Vert g-g_{k}\Vert_{L^{1}(I)}\le\frac{1}{k}$$ Such a $g_{k}$ can be expressed as $$x\mapsto g_{k}(x)=\sum_{i=0}^{N_{k}}g^{i}_{k}\mathbb{I}_{\{a_{i}\le x\le a_{i+1}\}}$$ where $g_{k}^{i}$ are real constants, $N_{k}\in\mathbb{N}$ and $a_{0}\le a_{2}\le\dots\le a_{N_{k}+1}$ where $I=(a_{0},a_{N_{k}+1})$. We have $$\int_{I}f_{\epsilon}(x)g_{k}(x)\text{d}x=\sum_{i=0}^{N_{k}}g_{k}^{i}\int_{a_{i}}^{a_{i+1}}f_{\epsilon}(x)\text{d}x$$
Let $[x]$ denote the integer part of $x$. We want to show that
$$\lim_{\epsilon\to 0}\int_{a_{i}}^{a_{i+1}}f_{\epsilon}(x)\text{d}x=(a_{i+1}-a_{i})\langle f\rangle$$
We obtain
\begin{align*} \int_{a_{i}}^{a_{i+1}}f_{\epsilon}(x)\text{d}x &=\int_{a_{i}}^{a_{i+1}}f\left(\frac{x}{\epsilon}\right)\text{d}x\\ &=\epsilon\int_{\tfrac{a_{i}}{\epsilon}}^{\tfrac{a_{i+1}}{\epsilon}}f(y)\text{d}y\\ &=\epsilon\left(\int_{\tfrac{a_{i}}{\epsilon}}^{\left[\tfrac{a_{i}}{\epsilon}+1\right]}f(y)\text{d}y+\int_{\left[\tfrac{a_{i}}{\epsilon}+1\right]}^{\left[\tfrac{a_{i+1}}{\epsilon}\right]}f(y)\text{d}y+\int_{\left[\tfrac{a_{i+1}}{\epsilon}\right]}^{\tfrac{a_{i+1}}{\epsilon}}f(y)\text{d}y\right) \end{align*}
Then, it is said that
$$\int_{\left[\tfrac{a_{i}}{\epsilon}+1\right]}^{\left[\tfrac{a_{i+1}}{\epsilon}\right]}f(y)\text{d}y=\langle f\rangle \left(\left[\tfrac{a_{i+1}}{\epsilon}\right]-\left[\tfrac{a_{i}}{\epsilon}+1\right]\right)$$
Second problem: where does this equality come from?
My guess is that $\langle\cdot\rangle$ depends on the context here and is not always the integral over $(0,1)$ or even $I$ and should be properly denoted by $$\langle f\rangle_{I}:=\frac{1}{\lambda(I)}\int_{I}f(x)\text{d}x$$ where $\lambda$ is the Lebesgue-measure. Still, I do not understand where does $I$ intervene (first problem above). I suppose that in the statement, it should have been written weak$^{\ast}$ convergence in $L^{\infty}(I)$ and the lemma should be $\lim\limits_{\epsilon\to 0}f_{\epsilon}=\langle f\rangle_{I}$ (and, in particular, for $I=(0,1)$, we have $\lim\limits_{\epsilon\to 0}f_{\epsilon}=\langle f\rangle:=\int_{0}^{1}f(x)\text{d}x$). Note that the notation $\lim f_{\epsilon}$ is confusing since it does hide the fact that it depends on the interval $I$ on which is considered the weak$^{\ast}$ convergence.
Could you confirm or infirm my guess, please?