Consider the family of functions
$$\left\{u_{\lambda}=|\sin (\lambda x)|\right\}_{\lambda>0}\subset L^{\infty}(\mathbb{R}) $$
This is a bounded subset of $L^{\infty}(\mathbb{R})$, with
$$\|u_{\lambda}\|_{\infty}\leq 1,\qquad \forall \lambda>0 $$
Our goal is to find a function $u\in L^{\infty}(\mathbb{R})$ so that
$$\lim_{\lambda \to +\infty}\int_{\mathbb{R}}fu_{\lambda}=\int_{\mathbb{R}}fu,\qquad \forall f\in L^1(\mathbb{R})\qquad (*) $$
(if you know some functional analysis, thanks to the Frechet-Riesz representation theorem $u$ represents the weak-star limit of the net $\left\{u_{\lambda}\right\}_{\lambda>0}$ as $\lambda\to +\infty$, which exists because $\left\{u_\lambda\right\}$ is bounded and hence weakly-star compact in $L^{\infty}(\mathbb{R})$ by the Banach-Alaoglu theorem).
Now, if the above limit holds for all $f$ in a dense subspace $D\subset L^1(\mathbb{R})$, then it holds for all $f\in L^1(\mathbb{R})$. Let me prove this. Let $f_k\to f$, with $f_k\in D$, be an approximating sequence. Then
\begin{align*}\left|\int fu_{\lambda}-\int fu\right|&\leq
\left|\int fu_{\lambda}-\int f_ku_{\lambda}\right|+\left|\int f_ku_{\lambda}-\int f_ku\right|+\left|\int f_ku-\int fu\right|\leq \\
&\leq \|f-f_k\|_1+\left|\int f_ku_{\lambda}-\int f_ku\right|+\|f_k-f\|_1\|u\|_{\infty}
\end{align*}
and all the three summands vanish by assumption.
The dense subspace $D$ we choose is the space of step functions, i.e. of the form
$$f=\sum_{i=1}^{N}\alpha_i\chi_{[a_i,b_i]} $$
where $\alpha_i\in \mathbb{R}$ and $\left\{[a_i,b_i]\right\}_{i=1}^{N}$ is a family of pairwise disjoint bounded intervals. It is not hard to show, using the fact that $\int_{n\pi}^{m\pi}|\sin x|dx=2(m-n)$ for all $m,n\in \mathbb{Z}$ and proceeding by approximation, we have
$$\lim_{\lambda \to +\infty}\int_{\mathbb{R}}\chi_{[a_i,b_i]}u_{\lambda}= \lim_{\lambda\to+\infty}\int_{a_i}^{b_i}|\sin (\lambda x)|dx=\lim_{\lambda\to+\infty}\frac{1}{\lambda}\int_{\lambda a_i}^{\lambda b_i}|\sin x|dx= \frac{2}{\pi}(b_i-a_i) $$
and therefore
$$\lim_{\lambda\to +\infty}\int_{\mathbb{R}}\chi_{[a_i,b_i]}u_{\lambda}=\int_{\mathbb{R}}\chi_{[a_i,b_i]}u$$ holds true when we choose
$$u=\frac{2}{\pi} $$
By linearity this extends to all step functions, and hence by the above density argument to all integrable functions. Therefore we may substitute $u=\frac{2}{\pi}$ in $(*)$ to obtain
$$\lim_{\lambda \to +\infty}\int_{\mathbb{R}}fu_{\lambda}=\frac{2}{\pi}\int_{\mathbb{R}}f,\qquad \forall f\in L^1(\mathbb{R})$$
EDIT: More in general, the above argument shows that if $g:\mathbb{R}\to \mathbb{R}$ is a bounded periodic function such that its integral mean over a period is $\alpha$, then
$$\lim_{\lambda\to +\infty}\int_{\mathbb{R}}f(x)g(\lambda x)dx=\alpha \int_{\mathbb{R}}f(x)dx,\qquad \forall f\in L^1(\mathbb{R}) $$
