Cocycle superrigidity Let $\Gamma$ be a group with a probability measure preserving action on $(X,\mu)$, and $H$ another group. Recall that a cocycle is a map $c:\Gamma\times X\to H$ such that $c(gg',x)=c(g,g'x)c(g',x)$. Two cocycles $c,c'$ are cohomologous if there is $f:X\to H$ such that $c(g,x)=f(gx)^{-1}c'(g,x)f(x)$.
Let $H$ be a discrete group. Say that $H$ is superrigid if for every lattice $\Gamma$ in a simple, higher rank Lie group $G$, and for every ergodic probability measure preserving action of $\Gamma$ on $(X,\mu)$, every measurable cocycle $c:\Gamma\times X\to H$ is cohomologous to a cocycle with values in a finite group.
For example, if $H$ is a hyperbolic group, then a result of Adams proves that $H$ is superrigid. 
My question is : Assume that $H$ contains a normal subgroup $H'$ of finite index, and that $H'$ is superrigid. Does it follow that $H$ is superrigid ?
 A: Yes. 
Instead of writing down explicitly such a cocycle (which is not hard), let me take this opportunity to explain how to think of such objects in a "cooridnate free" manner.
Given a cocycle $c:\Gamma\times X\to H$ you can consider the space $Y=X\times H$ and endow it with the product measure and with the $\Gamma\times H$ measure preserving action given by the right action of $H$ on the second coordinate and the twisted left action of $\Gamma$
$$(*)\quad\gamma\cdot(x,h)=(\gamma x,c(\gamma,x)h).$$
Clearly, $Y$ is $\Gamma\times H$-ergodic iff $X$ is $\Gamma$-ergodic.
Conversely, let $Y$ be an ergodic measure preserving $\Gamma\times H$-space on which $H$ acts freely and with a finite measured fundamental domain.
Then we may form the space of orbits $X=Y/H$ and endow it with a $\Gamma$-invariant finite measure, and any choice of a fundamental domain (ie a section $X\to Y$) will give us an identification $Y=X\times H$ such that the $\Gamma$-action on $Y$ will be realized as in $(*)$ for some cocycle $c$.
Different choices of fundamental domains will give cohomologous realization cocycles. In this cohomology class we could find a cocycle taking values in a subgroup $F<H$ iff we can find a $\Gamma$-invariant $H$-equivariant map $Y\to H/F$. It follows, in particular, that in this cohomology class we could find a cocycle taking values in a finite group iff $Y$ admits a $\Gamma$-invariant $H$-equivariant map $Y\to Z$, where $Z$ is a discrete space on which $H$ acts with finite stabilizers.
This is all standard.
We now answer the question in view of the above. We consider an ergodic measure preserving $\Gamma\times H$-space $Y$ on which $H$ acts freely and with a finite measured fundamental domain.
Note that $Y$ need not $\Gamma\times H'$-ergodic, but the finite group $H/H'$ acts transitively on the collection of $\Gamma\times H'$ ergodic components.
Let $Y=\cup_{i=1}^n Y_i$ be the corresponding ergodic decomposition.
Note that each $Y_i$ is an ergodic measure preserving $\Gamma\times H'$-space on which $H'$ acts freely and with a finite measured fundamental domain.
By the assumption on $H'$ (which I object calling "super-rigidity") we can find for each $i$ a $\Gamma$-invariant $H'$-equivariant map $\pi_i:Y_i\to Z_i$, where $Z_i$ is a discrete space on which $H'$ acts with finite stabilizers.
Set $Z=\cup_{i=1}^n Z_i$ and consider the $\Gamma$-invariant $H'$-equivariant map $\pi=\cup \pi_i:Y\to Z$.
Consider now the induction of $\pi$ from $H'$ to $H$, namely the map $H\times_{H'} Y \to H\times_{H'} Z$. As $Y$ is an $H$ space we may identify $H\times_{H'} Y$ with $H/H'\times Y$ in a $H\times \Gamma$-equivariant way.
We obtain a $\Gamma$-invariant $H$-equivariant map $Y \to \text{Maps}(H/H',H\times_{H'} Z)$. We are done, observing that $\text{Maps}(H/H',H\times_{H'} Z)$ is a discrete space on which $H$ acts with finite stabilizers.
