Are slices of Borel sets also Borel sets? 
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*Suppose E is Borel in $\mathbb{R}^{a+b}$. Show that the slice $E^{x_1}$={$x_2\in\mathbb{R}^b|(x_1,x_2)\in E$} is Borel for each $x_1\in \mathbb{R}^a$.

What I have so far:
First, any sigma algebra containing the open sets contains every Borel set. Also, I know that given an open subset of $\mathbb{R}^{a+b}$, then for every $x_1\in\mathbb{R}^a$, $E^{x_1}$={$x_2\in\mathbb{R}^b|(x_1,x_2)\in E$} is also open. This implies that $E\in F$ and is Borel. I'm not sure if this reasoning is enough.


*Show that this statement is not true if we replace both instances of "Borel" with "Lebesgue".

I thought I could use the Banach-Tarski Paradox. The unit ball is divided into 5 non-lebesgue measurable sections. However, take one of these sections and name it $A\in\mathbb{R}^3$. $A\times${$0$} is Lebesgue measurable (0 measure).
 A: Here are some hints:
To see why $E^{x_1}$ is borel, consider the (obviously borel) function
$f(x_2) = (x_1,x_2)$. Do you see why $E^{x_1}$ is the preimage of a borel set by a borel function?
Remember that the lebesgue measurable sets are exactly
$\{ \text{borel sets} \} \cup \{ \text{extra null sets} \}$, so we will probably have to use these extra null sets somehow.
More concretely: Take your favorite (lebesgue) nonmeasurable set $X$. Can you show that $X \times \{0\}$ is (lebesgue) measurable in $\mathbb{R}^2$? Remember you'll probably have to use nullsets!
Once you've done this, $(X \times \{0\})^0$ is obviously $X$, which was nonmeasurable by assumption.

I hope this helps ^_^
A: Consider $\{A\subseteq \mathbb{R}^{a+b}:A^{x_1} \text{ is Borel in }\mathbb{R}^b\}$. This set contains open sets of $\mathbb{R}^{a+b}$ (because projection is open) and is stable under countable unions and complements, so it consists of the Borel $\sigma$-algebra on $\mathbb{R}^{a+b}$. The same technique is quite useful in other situations where you can exhibit that some class of sets behaves nicely, hence the $\sigma$-algebra generated by them also behaves nicely.
When it comes to Lebesgue measures, we have to include null sets. Take for example, the Vitali set $V\subseteq \mathbb{R}$. Then $V\times \{0\}$ is a null set in $\mathbb{R}^2$, hence Lebesgue measurable but its projection is not.
