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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 $X$

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If the set $A\times G$ is a rectangle, $A$ must be a measurable section. –  Michael Greinecker Oct 12 '12 at 21:27
If $A \times G$ is a rectangle then it is obvious that A is a borel set. –  user27456 Oct 12 '12 at 21:38
Otherwise, the set will nly be analytic, not Borel. –  Michael Greinecker Oct 12 '12 at 22:30

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

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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 $G$.

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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 .$

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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.

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I do not understand what you are saying. The reals are an uncountable union of singletons, and yet they are Borel. There is more than one way of writing a set as a union of other sets. –  Andres Caicedo Oct 12 '12 at 22:01
I think you are right in the following sense: a property of a $\sigma$-algebra is that the countable union of some of its sets is still a set on the $\sigma$-algebra, meaning that the uncountable one is not necessarily a set on the $\sigma$-algebra (but still able to be, like in your example). But, if we actually ask this uncountable union to be not in the $\sigma$-algebra, the result seems valid to me. Maybe it is simpler if we drop this condition and just set a non-Borel set with no more properties whatsoever, then try to discuss measurability matters. –  busman Oct 12 '12 at 22:08
What does WLOG? –  user27456 Oct 12 '12 at 22:20
Without (any) loss of generality ("sem (ninhuma) perdida de generalidade"). –  busman Oct 12 '12 at 22:26
@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. –  Rafa Oct 13 '12 at 14:14

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