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 Sep 13 comment Intersection of Borel subset with line If you prove that (in your first comment), you prove that each Borel set of $\mathbb{R}^2$ is of type $A(y)$ for some Borel set $A$ and $y\in \mathbb{R}$. That's not the same as proving that each $A(y)$ is Borel. The point of my proof is to show that every Borel set $A$ is an element of $M$. Then by definition of $M$, it follows that $A(y)$ is Borel for all $y\in\mathbb{R}$. Here is another proof: Given $A$ Borel and $y\in\mathbb{R}$ consider the function $x\mapsto (x,y)\in\mathbb{R}^2$. It is continuous. Then if you know that the preimage of a Borel set is Borel, you are also done. Sep 13 comment Sum from $i$ to $\lg(n)$ @sonic: You are welcome. marty:The $t_n$ are only defined for $n$ a power of 2. Sep 13 suggested rejected edit on Sum from $i$ to $\lg(n)$ Sep 13 answered Sum from $i$ to $\lg(n)$ Sep 13 comment Nested interval question involving limits Assume that $a0$ so that $a+\varepsilon < c$. Can you use the definition of convergence to obtain a contradiction? Sep 13 comment Can I determine the groups and homomorphisms? What are your ideas? The first and last lines seem pretty easy to figure out. Sep 13 comment Intersection of Borel subset with line Actually, now I think the version with starting with open cubes is the easiest. The thing about the title is that $A(y)$ is not actually intersection with a line (that would be trivially Borel), but intersection followed by projection to one of the coordinates, which is definitely more subtle. This kind of measurability result is needed for the statement of the Fubini Theorem. Sep 13 revised Intersection of Borel subset with line deleted 2 characters in body Sep 13 comment $\overline{E}\subset \bigcup_{x\in E}\overline{B_x(r_x)}$ if $E\subset \bigcup_{x\in E}\overline{B_x(r_x)}$ In that case $\overline E = [0,1]$ is covered by the union if we choose the $r_x$ for $x\in (0,1)$ in the same way as above. But we could also take an enumeration of $E\cap (0,1) = \{ x_1,x_2,x_3,.. \}$ and let $r_{x_i}' = \frac{r_{x_i}}{2^{i+1}}$ as well as $r_{0} = r_{1} = \frac{1}{8}$ Then the union of the $\overline {B_{x}(r_x')}$ has measure at most $\frac{3}{4}$ and thus cannot cover $[0,1]$. Sep 13 answered $\overline{E}\subset \bigcup_{x\in E}\overline{B_x(r_x)}$ if $E\subset \bigcup_{x\in E}\overline{B_x(r_x)}$ Sep 13 comment Intersection of Borel subset with line Absolutely. Thanks for telling me, I corrected it. Sep 13 revised Intersection of Borel subset with line edited body Sep 13 comment Intersection of Borel subset with line Right, it is a bit easier to start with closed sets rather than with open for the proof, so I changed that. Sep 13 revised Intersection of Borel subset with line added 2 characters in body Sep 13 comment Intersection of Borel subset with line @Carl Mummert: Please have a look at the question again. It isn't asking about the projection of the Borel set, but about the sets $A(y)$ for various $y\in\mathbb{R}$, there is a difference. Sep 13 answered Intersection of Borel subset with line Sep 12 revised 1. Measurable sets : uncountable union, 2. null set is disjoint from translated rationals added 349 characters in body Sep 12 answered 1. Measurable sets : uncountable union, 2. null set is disjoint from translated rationals Sep 12 awarded Custodian Sep 12 reviewed Reviewed Proving a Square Root of a Symmetric Matrix