0
$\begingroup$

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 $X$

  • 0
    Otherwise, the set will nly be analytic, not Borel.2012-10-12

4 Answers 4

1

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 $A = f^{-1}(A \times G)$.

0

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 $G$.

0

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 .$

-2

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.

  • 1
    @Busman, you can in fact construct non-Borel sets as an uncountable reunion of Borel sets, but that doesn't means any such reunion is. You usually need more information to be able to proof that it's non Borel. In fact, you would rather need (thanks to Suslin's theorem) to proof that its complement is not anaylitic, i.e. the reunion over a Polish set of Borel Sets.2012-10-13