Is $A$ a Borel set?

Question

Let be $X$ a metric compact space and $(G,+)$ a topological compact abelian group. Let be $\mathcal{A}$ the Borel $\sigma$-algebra of $X$ and $\mathcal{B}$ the Borel $\sigma$-algebra of $G$. Consider in $X\times G$ the product $\sigma$-algebra.

My Question: I have a Borelian $A\times G$ of $X\times G.$ I want to show is that the set $ A $ must be necessarily a borel set of $X$

Answer 1

Suppose that $A$ was not a Borel set. We may assume, WLOG, that A is the non-countable reunion of some $A_i \subset X,\ i\in I$ Borel sets. Then, $$A \times G = ( \bigcup_{i \in I} A_i ) \times G = \bigcup_{i \in I} (A_i \times G).$$ But if that's so, then $A \times G$ is not a Borel set, so we get a contradiction and the result follows.

Answer 2

We will show by induction that if $A\times G$ is $\mathbf\Sigma^0_\xi$, then $A$ is $\mathbf\Sigma^0_\xi$.

The first case, is simply the fact that the projection is an open map.

Then, first note that if it were true for $\mathbf\Sigma^0_\xi$, then it is true for $\mathbf\Pi_\xi^0$. This is thanks to the fact that $(A\times G)^C=A^C\times G$.

Then, the general case results in the fact that the projection (in fact, any function) $\pi$ satisfies $\pi(\cup_{i\in I}A_i)=\cup_{i\in I} \pi(A_i)$.

Note that the you don't really need any hypothesis over $G$ other than to properly define the Borel sets. On the other side, it's essential, for the $\mathbf\Pi_\xi^0$ case that your set is a rectangle with side $G$.

Answer 3

The solution is actually quite simple. Just follow the suggestion of Michael Greinecker, and note that if $ A \times G $ is a borel set, there is no other option for $ A ,$ is easy to see that a must be a measurable section of $ A \times G $, this is due to the fact that $ G $ is not any borel set, but the whole space, in this case the projection of any measurable section of $ A \times G $ coincides with $ A .$

Answer 4

Fix any element $y_0$ of $G$ (the identity, for instance) and define $f : X \to X \times G$ by $f(x) = (x,y_0)$. $f$ is clearly Borel ($\pi_X \circ f = id_X$ and $\pi_G \circ f$ is a constant map) and $A = f^{-1}(A \times G)$.