@Simon: I will take a look on the book you referred - so maybe you can put an extended reference (which chapter/page?). Before looking into this: is there significant difference which does not allow to repeat the proof of strong convergence by Karatzas and Shreve for the multidimensional case?

@Asaf: I mean that they are arbitrarily fixed prior to the construction. About your previous example: you made an unjustified assumption $G = A\times A$.

@Asaf Karagila: $G$ and $A$ are given and we can do nothing with the choice of them.

@Condor: thank you, but it's easy to provide a counterexample.

@Clinton Conley: I've edited - I am looking for invariant subsets of given compacts. And $\bar{G}$ means the closure.

What if to write it with respect to independent Brownian motions?

Thank you for these nice comment about empty set, but I need compactness to say also that the set $A$ is non-empty, it is important. Could we make the proof smaller? Say, for all $x\in A$ and all $n$ we have $K(x,A_n) = 1$ (here I use $K(x,A_n)$ rather then $\phi(x,A_n)$ to stress that it is a measure). I think that by continuity of measure we have $$ K(x,A) = \lim\limits_{n\to\infty}K(x,A_n) = 1 $$ because the sequence $A_n$ is non-increasing and $A$ is an intersection of all these sets.