Ergodic action of a subgroup Are there any examples of $H < G$ such that for any pmp ergodic action of the group $G$ on a standard proba space $(X,\mu)$ there exists a set $A$ of $\mu(A)>C$ such that the  action of the subgroup $H$ on $A$ is ergodic? It seems that this can happen only in finite index case...
Conclusion:
The combination of the answers of Alex and Clinton give the complete answer to my question.
 A: You can take $G$ to be any simple Lie group with finite center, and $H$ to be any non-compact Lie subgroup. Then, if the action of $G$ is ergodic, so is the action of $H$. This statement is called the "Moore ergodicity theorem". 
In fact, both the actions of $G$ and $H$ will be automatically mixing. This follows from the theory of unitary representations of $G$. 
A: We seem to be talking past each other in the limited space provided by the comments, so maybe I can express myself better in the room provided by the answer box.  You indicated that you were focused on countable discrete groups.  For countable discrete $G$ and $H < G$ the following are equivalent.


*

*There exists a constant $c > 0$ such that for any measure-preserving ergodic action of $G$ on a standard probability space $(X,\mu)$ there is a measurable $H$-invariant subset $A \subseteq X$ on which $H$ acts ergodically and with $\mu(A) \geq c$.

*$H$ has finite index in $G$.
(not 2) $\Rightarrow$ (not 1).  If $H$ has infinite index, consider the ergodic action of $G$ by shifts on $[0,1]^{G/H}$ with the usual product measure.  $H$ doesn't act ergodically on any set of positive measure, as the components of its ergodic decomposition are null.  More concretely, if $A \subseteq [0,1]^{G/H}$ is an $H$-invariant set of positive measure, there are disjoint sets $B,C \subseteq [0,1]$ such that a positive measure of elements of $A$ send the coset $H$ to something in $B$, and the same for $C$. Since the $H$ action doesn't shift this coset, this shows $H$ doesn't act ergodically on any set of positive measure.  (A better way of writing this argument is simply that $f \mapsto f(H)$ is a null-to-one $H$-invariant Borel function from $[0,1]^{G/H}$ to $[0,1]$, which is enough to preclude ergodicity of the $H$ action on any non-null set.)  Thanks to Robin for fixing the error in the original answer.
2 $\Rightarrow$ 1.  Say $H$ has index $n$ in $G$, and fix coset representatives $g_1, \ldots, g_n$.  Fix a measure-preserving ergodic action of $G$ on $(X, \mu)$.  Suppose that $B \subseteq X$ has positive measure and is $H$-invariant.  Then $\bigcup_{i \leq n} (g_i \cdot B)$ is $G$-invariant and $\mu$-positive, so by ergodicity has measure $1$.  This implies that $\mu(B) \geq 1/n$.  So picking $A$ to be an $H$-invariant set of smallest positive measure, $H$ acts ergodically on $A$ and $\mu(A) \geq 1/n$.  Thus $c = 1/n$ works for all actions.
A: Edit: Misread the question.
Let $X = \lbrace 0,1 \rbrace^\mathbb{Z}$ with the $(\tfrac{1}{2},\tfrac{1}{2})$ Bernoulli measure and let $T$ be the shift map $(T\omega)(n) = \omega(n+1)$. Put $G = \mathbb{Z}^2$ and define a $G$-action $S$ on $X$ by $S^{(n,m)} = T^{n + m}$. Take $A = X$ and $H = \mathbb{Z} \times \lbrace 0 \rbrace$. Since $T$ is ergodic both $S$ and $S$ restricted to $H$ are ergodic.
