First, we note that there is a natural bijection ${\cal P}(\omega) \to \{0,1\}^\omega$ and endow the latter with the product topology (where $\{0,1\}$ carries the discrete topology). So we get a compact space that [can be endowed with a Haar measure](https://en.wikipedia.org/wiki/Haar_measure).


Is there a concrete definition of that measure?


More concretely, let $A\_0 = \{a \in {\cal P}(\omega): 0 \in a\}$; so what is the Haar measure of $A\_0$? (I would expect $1/2$, as "half of the subsets of $\omega$ contain $0$ as a member", but this is just heavy handwaving.)



Haar measure on ${\cal P}(\omega)$
As Arno said, for "Haar measure" you need a group operation.  

On $\{0,1\}$, we can define a group using "addition modulo 2". Normalized Haar measure for that group has each point of measure $1/2$.


Then on $\{0,1\}^\omega$ we use the product group. In that case, Haar measure is the product measure. $A\_0$ depends only on the first coordinate, and we get measure of $A\_0$ in $\{0,1\}^\omega$ is equal to the measure of $\{0\}$ in $\{0,1\}$. So the measure of $A\_0$ is $1/2$.


Of course if you use some different group operation on $\{0,1\}^\omega$, you may get some different Haar measure.


