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.



Is a bounded sequence of $H^1(\Omega)$ tight?
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.


