Is a bounded sequence of $H^1(\Omega)$ tight? Assume $\Omega$ is  a bounded subset of $\Bbb R^d$ and  $ (u_n)_n$ is a bounded sequence of the Sobolev space $H^1(\Omega)$. 
Question: Can  we say that $ (u_n)_n$  is tight in $L^2(\Omega)$ namely: For very $\varepsilon>0$ there exists a compact set $K_\varepsilon\subset \Omega$ such that $$ \sup_{n}\int_{\Omega\setminus K_\varepsilon }|u_n(x)|^2dx<\varepsilon$$
I failed to find a counter example. Any help or reference in which I can find related topic is welcome.
 A: The answer is yes if your domain has compact embedding $H^1(\Omega)\to L^2(\Omega)$ (that is, for any regular enough domain), or you consider $H^1_0(\Omega)$, but no in general. Counterexample is as follows:
Let $\Omega = (\frac{1}{2}, 1)\cup (\frac{1}{8}, \frac{1}{4})\cup (\frac{1}{32}, \frac{1}{16})\cup \ldots = I_1\cup I_2\cup \ldots \subset \mathbb{R}$ and consider functions $f_n = \frac{1}{\sqrt{|I_n|}}\chi_{I_n}$. They all are bounded in $H^1(\Omega)$ (indeed, they has norm $1$), but you can't find a compact $K\subset \Omega$ such that $K\cap I_n \neq \varnothing$ for all $n$ because such a $K$ has to contain $0$ which is not in $\Omega$.
If you do not like nonconnected examples, consider $\Omega_1 = \Omega \times [\frac{1}{2}, 1) \cup (0, 1)\times (0, \frac{1}{2})$ (it looks like a comb), choose a smooth function $g$ with support in $(\frac{3}{5}, \frac{4}{5})$ and consider $g_n(x, y) = f_n(x)g(y)$.
On the other hand, if you have a compact embedding then the answer is yes. First of all let us consider a sequence of compact sets $K_n\subset \Omega$ such that $K_n \subset Int(K_{n+1})$ and $\bigcup K_n = \Omega$ (such a sequence of compacts exists for any manifold and a fortiori for any open subset of $\mathbb{R}^d$). Assume that there exists $\varepsilon > 0$ such that for any $n$ there exists $u_n\in H^1(\Omega)$, $||u_n||_{H^1(\Omega)} = 1$ such that $||u_n||_{L^2(\Omega \backslash K_n)} \ge \varepsilon$. By compactness of embedding (passing to a subsequence, but for notational simplicity I will assume that it is the same sequence) we may assume that $u_n\to u$ in $L^2(\Omega)$. But functions $v_n = |u|^2 \chi_{\Omega \backslash K_n}$ has a common $L^1$-majorant $|u|^2$ and converges pointwise to $0$. Therefore $||v_n||_{L^2(\Omega)} < \varepsilon$ for some $n$ by Lebesgue Theorem. On the other hand $w_m = u_m \chi_{\Omega \backslash K_n}$ converges to $v_n$ in $L^2(\Omega)$. Thus, for big enough $m$ $||w_m||_{L^2(\Omega)} < \varepsilon$. But on the other hand $||w_m||_{L^2(\Omega)} \ge ||u_m||_{L^2(\Omega \backslash K_m)}\ge \varepsilon$ for $m > n$ -- a contradiction.
A: Aleksei already answered, but here is another way to see it. Since $\Omega$ is bounded, any bound in the $L^q(\Omega)$ norm leads tightness in the $L^p(\Omega)$ norms with $p<q$ : this is a consequence of Hölder's inequality
\begin{align*}
\|\cdot\|_{L^p(\Omega\setminus K)} \leq |\Omega\setminus K|^{\frac{1}{p}-\frac{1}{q}} \|\cdot\|_{L^q(\Omega\setminus K)}.
\end{align*}
Now, if indeed your domain is not too ugly so that you have the embedding $H^1(\Omega)\hookrightarrow L^{2^\star}(\Omega)$, you have your tightness property because $2^\star>2$ (with the appropriate changes if $d=1$ or $d=2$).
