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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 $a<c$. Then choose $\varepsilon >0$ 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