I like the sandwhich approach better than mine.
Thanks for the pointers. I really lost the forest from the trees on this problem.
I got it stuck in my head that perhaps the irrationals were "dense enough" to create 2 subsets with the same measure. It didn't dawn on me to use L.measure...
Yep. I messed that up too. One more time.
By DeMorgan, C\left(\cap_{k=1}^{\infty} E_k\right)=C\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]=\bigcup_{k=1}^{\infty}E^{c}_{k}
Consider: m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)...
By DeMorgan, C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}
A lower bound immediately comes to mind. I'm still stewing on how to get an upper bound to pop out.
Lower Bound:
Consider: m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]...
I answered your question the best I could.
Your question is closely related to the question I am posing. If I could simply answer it, then I wouldn't've started the thread.
If I can construct a counterexample of 2 disjoint and noncountable subsets on [0,1] of measure 1, then the measure of the complement of intersections would be 1. However, I have not been able to find such a counterexample. The answer based on what I have considered is 0.
Homework Statement
Let \left\{E_{k}\right\}_{k\in N} be a sequence of measurable subsets of [0,1] satisfying m\left(E_{k}\right)=1. Then m\left(\bigcap^{\infty}_{k=1}E_{k}\right)=1.
Homework Equations
m denotes the Lebesgue measure.
"Measurable" is short for Lebesgue-measurable.
The Attempt...
I was hoping that I was missing a more strategic approach. (I had already tried working through it with q=3 but ran into unboundedness of g. Working with 2 solved the unboundedness problem, but missed the mark.)
Problem does solve with p=3 and q=3/2. Thanks for the help.
Homework Statement
Show that: \left(\int^{0}_{1}\frac{x^{\frac{1}{2}}dx}{(1-x)^{\frac{1}{3}}}\right)^{3}\leq\frac{8}{5}
Homework Equations
Holder inequality.
The Attempt at a Solution
First, I took the cube root of each side. This let me just deal with the 1-norm on the left. Then I...
Here's what I am thinking.
Consider: \int_{[0,1]}f(y)\left[\int_{[0,1]}\frac{1}{|x-y|^{1/2}}dx\right]dy=\int_{[0,1]}f(y)\left[2\left(\sqrt{1-y}-\sqrt{y}\right)\right]dy\leq\int_{[0,1]}f(y)\cdot 2<∞. Therefore \int_{[0,1]^{2}}\frac{f(y)}{|x-y|^{1/2}}<\infty by Tonelli's Theorem. Then...
Homework Statement
Let f:[0,1]→ℝ be non-negative and integrable. Prove that \int_{[0,1]}\frac{f(y)}{|x-y|^{1/2}}dy is finite for ae x in [0,1]
Homework Equations
This looks like a Fubini/Tonelli's Theorem problem from the problem givens.
The Attempt at a Solution
I honestly don't know...
The definition the text gave for an Rn interval was the cross product of: av ≤ xv ≤ bv (v= 1, 2, ..., n). It acknowledged other intervals (open, semi open), but stated that intervals should be assumed to be closed unless specifically mentioned.
Yeah. I noticed those problems when I went back to the problem (again). Here's where I am with the problem:
Let I be the collection of all intervals of Rn and ƩI be the σ-algebra generated by I. Let G be the set of all open subsets of Rn and ƩG be the σ-algebra generated by G.
Pick an...
I got confirmation from the professor that he uses the word "coincide" to mean "equal". So both directions of containment need to be proved. I am also changing the notation from Ʃ(x) to Ʃx to avoid confusion that Ʃ is a function.
This was my first attempt to show \Sigma_{G}\subset\Sigma_{I}...
Thanks for confirming this. I spoke to the professor about this. His comment was "My native language is Russian. Questions clarifying how I use articles are completely fair."
Is there only 1 σ-algebra generated for a set?
Consider M={1,2}. Ʃ(M)={∅,M} satisfies the definition of a σ-algebra. However, Ʃ(M)={∅,{1},{2},{1,2}} also satisfies the definition of a σ-algebra. However, the way that my text presents these problems (Prove that the σ-algebra generated...)...