Proving Measurability of ##A## from ##E=A \cup B## with ##|B|=0##

mahler1
Messages
217
Reaction score
0

Homework Statement



Let ##E \subset \mathbb R^n## be a measurable set such that ##E=A \cup B## with ##|B|=0## (##B## is a null set). Show that ##A## is measurable.

The Attempt at a Solution



I know that given ##\epsilon##, there exists a ##\sigma##-elementary set ##H## such that ##E \subset H## and ##m_e(H-E)<\epsilon##. How can I construct a ##\sigma-##elementary set ##H'## such that ##m_e(H-A)<\epsilon##?. Any suggestions would be appreciated
 
Last edited:
Physics news on Phys.org
What does |B| mean? Is it the measure of B or is the cardinality of B (in which case B is empty so that E= A).

In your last sentence do you not mean "How can I construct an elementary set H' such that m_e(H&#039;- A)&lt;\epsilon"?
 
  • Like
Likes 1 person
HallsofIvy said:
What does |B| mean? Is it the measure of B or is the cardinality of B (in which case B is empty so that E= A).

In your last sentence do you not mean "How can I construct an elementary set H' such that m_e(H&#039;- A)&lt;\epsilon"?

Edited, thanks for the correction.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top