Proving Caratheodory's Condition for Measurable Sets in Measure Theory

  • Thread starter Thread starter SqueeSpleen
  • Start date Start date
  • Tags Tags
    Condition
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
SqueeSpleen
Messages
138
Reaction score
5
Measure Theory, Caratheodory condition

The set [itex]E \subset ℝ^{p}[/itex] satisfy Caratheodory's condition if:
[itex]\forall A \subset ℝ^{p}[/itex]
[itex]m_e (A) = m_e(A \cap E) + m_e(A \cap E^c)[/itex]
Prove that if E is measurable then E satisfy the Caratheodory's condition.
I know
[itex]m_e (A) \leq m_e(A \cap E) + m_e(A \cap E^c)[/itex] for the subaditivity of the external measure, but I didn't found a way to prove:
[itex]m_e (A) \geq m_e(A \cap E) + m_e(A \cap E^c)[/itex]

I'll post again if I figure out something, but I already skipped this some days ago because I was an entire hour without any advance at all.
Feel free to make any correction's, it's the first time I write something about this subject in English and I'm not sure if I'm chosing the correct words.
 
Last edited:
Physics news on Phys.org
SqueeSpleen said:
The set [itex]E \subset ℝ^{p}[/itex] satisfy Caratheodory's condition if:
[itex]\forall A \subset ℝ^{p}[/itex]
[itex]m_e (A) = m_e(A \cap E) + m_e(A \cap E^c)[/itex]
Prove that if E is measurable then E satisfy the Caratheodory's condition.
I know
[itex]m_e (A) \leq m_e(A \cap E) + m_e(A \cap E^c)[/itex] for the subaditivity of the external measure, but I didn't found a way to prove:
[itex]m_e (A) \geq m_e(A \cap E) + m_e(A \cap E^c)[/itex]

I'll post again if I figure out something, but I already skipped this some days ago because I was an entire hour without any advance at all.
Feel free to make any correction's, it's the first time I write something about this subject in English and I'm not sure if I'm chosing the correct words.

Some books present the Caratheodory condition as the definition of measurability of E. What definition does your book use?
 
A set is measurable if [itex]\forall \epsilon > 0[/itex] [itex]\exists U \in \cal{U}[/itex] such that:
[itex]A \subset U[/itex] and [itex]m_e (U-A) < \epsilon[/itex]
[itex]\cal{U}[/itex] is the family of the [itex]\sigma[/itex]-elemental sets.
I also know that a set E is measurable then [itex]m_i(E)=m_e(E)[/itex]
 
Last edited:
On a related topic (I don't know if I'd better create a new thread or not).

I have to find a decreasing sequense [itex]\left \{ E_k \right \}_{k \in \mathbb{N}} \subset P(R^p)[/itex] such that [itex]m_e(\displaystyle\bigcap\limits_{i=0}^{\infty} E_k) < \lim_{k \rightarrow \infty}m_e(E_k)[/itex]
I think that the following sequence would work:
Notation: [itex]\overline{B}(c,r)=\left [ c-r,c+r \right ][/itex]
[itex]E_1 = \overline{B}(0,1)[/itex]
[itex]E_2 = \overline{B}(0,\frac{1}{2}) \cup \overline{B}(1,\frac{1}{2})[/itex]
...
[itex]E_n = \displaystyle\bigcup\limits_{i=0}^{n-1} \overline{B}(\frac{k}{n},\frac{1}{n})[/itex]
What I'm trying is to get a set which intersection is countable then of null measure.
While the limit of its measures is 1.
It's this idea right or I'd better to search other way of solving this? (If feel like it's not, I think that the intersection will not be countable, it will be [0,1] but I liked this idea so I needed to write it clearly to see that's not good).
I'd try to do it with a no measurable set?
PD: The correct term is "outer measure" not "exterior measure"? Or both are right?
 
Last edited: