Criteria for Compactness of a Closed in $L^2$ Spaces $(X, \mathcal{B}, \mu)$ is a measure space.


*

*Is there any well-known criteria for compactness of a closed set in $L^2(X, \mu)$?

*If the answer is negative what about $L^2(\mathbb{R}^n,\mu)$(in this part $\mu$ is Lebesgue measure)?
 A: For Lebesgue measure, look at the Fréchet-Kolmogorov theorem.
For a general measure space, see Theorem 4.7.28 of Bogachev's Measure Theory, which is attributed to Riesz.  Let $\pi = \{E_1, \dots, E_k\}$ be a collection of disjoint measurable subsets of $X$ having finite positive measure.  ($\pi$ is almost a partition of $X$ but it can omit a set of cofinite measure.)  For $f \in L^p(X,\mu)$, let $$\mathbb{E}^\pi f(x) = \begin{cases} \frac{1}{\mu(E_i)} \int_{E_i} f\,d\mu, & x \in E_i \\ 0, & x \notin \bigcup E_i.\end{cases}$$ Note that all the integrals exist since they are taken over sets of finite measure.
Write $\pi_1 \le \pi_2$ if every set in $\pi_1$ is a union of sets from $\pi_2$, up to sets of measure zero.  Think of $\pi_2$ as being a sort of "refinement" of $\pi_1$, except that $\pi_2$ can also pick up some of $X \setminus \bigcup \pi_1$.  
Then the result is that $K \subset L^p(\mu)$ is compact iff it is closed and bounded and $\mathbb{E}^\pi f \to f$ in $L^p$ norm, uniformly on $K$, as $\pi$ becomes increasingly refined.  More precisely, a closed bounded $K$ is compact iff  for every $\epsilon$ there exists $\pi_0$ such that for all $\pi \ge \pi_0$ and all $f \in K$ we have $\|\mathbb{E}^\pi f - f\|_{L^p(\mu)} < \epsilon$.  
(Bogachev states it in the following equivalent way: view $\{\sup_{f \in K} \|\mathbb{E}^\pi f - f\|_{L^p}\}$ as a net in $[0,\infty)$ indexed by the set of all $\pi$.  Then the necessary and sufficient condition is that this net should converge to 0.)
