It is well known by the Sobolev embedding theorem, that the embedding $$ W^{1,p}(\Omega)\to L^r(\Omega) $$ is compact for every $1<r<p^*=\frac{np}{n-p}$ where $1<p<n$. Here $\Omega$ is a bounded and smooth domain in $\mathbb{R}^n$. Let $1<p<n,\,1<r<p^*$. Then if $\{u_n\}$ is a bounded sequence in $W^{1,p}(\Omega)$, then by the above compact embedding, there exists $u\in L^r(\Omega)$ such that $$ u_n\to u\text{ weakly in }W^{1,p}(\Omega),\quad u_n\to u\text{ strongly in }L^r(\Omega) $$ and there exists $g\in L^r(\Omega)$ such that $|u_n|\leq g$ a.e. in $\Omega$ and $\nabla u_n\to \nabla u$ weakly in $L^r(\Omega)$.
I understand the existence of $u$ with the above properties (follows by the above compact embedding), but I am unable to understand how does such $g$ exists, i.e. there exists $g\in L^r(\Omega)$ such that $|u_n|\leq g$ a.e. in $\Omega$.
Can someone please explain it or give some suitable reference to this. Thank you in advance.
First note that the weak convergence of the gradient should be in $L^p$, not in $L^r$ (the gradient doesn't enjoy higher integrability).
As for the dominating function, start with $u_{n_1}=u_1$, and then pick $n_2>n_1$ such that $\| u-u_{n_2}\|_{L^r} \leq 2^{-1}$. Inductively, once we've chosen $n_k$, choose $n_{k+1}>n_k$ with $\| u-u_{n_{k+1}}\|_{L^r}\leq 2^{-k}$. Consider now $g:=|u_{n_1}|+\sum_{k=1}^\infty |u_{n_{k+1}}-u_{n_k}|$. By the triangle inequality we have $$ \| g\|_{L^r} \leq \|u_{n_1}\|_{L^r} + \sum_{k=1}^\infty (\| u_{n_{k+1}}-u\|_{L^r} + \| u-u_{n_k}\|_{L^r})<\infty. $$ Moreover, for any fixed $K$, by the reverse triangle inequality we have $$ g\geq |u_{n_1}|+ \sum_{k=1}^K |u_{n_{k+1}}-u_{n_k}|\geq |u_{n_1}|+ \sum_{k=1}^K |u_{n_{k+1}}|-|u_{n_k}| =|u_{n_{K+1}}|. $$ Therefore, $g$ dominates your subsequence pointwise.
You can also look for this result in Brezis' book on functional analysis, in the section on $L^p$ spaces.