Is A_epsilon Non-Empty and Measurable in Lebesgue's Theory?

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Can you use this information to show that $A _ {\epsilon}$ has empty interior?In summary, we are given an enumeration of rational numbers in the interval $[0,1]$ and a set $A_{\epsilon}$ defined as the complement of the union of intervals $I_n$. We need to show that $A_{\epsilon}$ is measurable, non-empty, and has empty interior. Additionally, we need to show that the measure of the set $B$ defined as the union of sets $A_{\frac{1}{k}}$ is equal to 1.To prove that $A_{\epsilon}$ has empty interior, we can use the fact that its complement contains every rational point in the interval.
  • #1
fsblajinha
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Is $q_ {1} q_ {2}, ...,$ an enumeration of the rational $[0,1]$. For every $\epsilon> 0$, let $\displaystyle{A _ {\epsilon}: = [0,1]-\bigcup_ {n \geq 1} I_{n}}$, where $I_{n}:= [q_{n} - \frac {\epsilon}{2^{n+1}} , q_{n}\frac{\epsilon}{2^{n + 1}}]\cap [0,1]$.

Show that:

1 - $A _ {\epsilon}$ is measurable, non-empty and empty inside;

2 - $\lambda^{*}(B)=1$, where $\displaystyle B:=\bigcup_{k\geq 1}{A_\frac{1}{k}}$
 
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  • #2
fsblajinha said:
Is $q_ {1} q_ {2}, ...,$ an enumeration of the rational $[0,1]$. For every $\epsilon> 0$, let $\displaystyle{A _ {\epsilon}: = [0,1]-\bigcup_ {n \geq 1} I_{n}}$, where $I_{n}:= [q_{n} - \frac {\epsilon}{2^{n+1}} , q_{n}\frac{\epsilon}{2^{n + 1}}]\cap [0,1]$.

Show that:

1 - $A _ {\epsilon}$ is measurable, non-empty and empty inside;

2 - $\lambda^{*}(B)=1$, where $\displaystyle B:=\bigcup_{k\geq 1}{A_\frac{1}{k}}$
Hi fsblajinha, and welcome to MHB!

Can you tell us what progress you have made with this problem, and where you need assistance?
 
  • #3
Opalg said:
Hi fsblajinha, and welcome to MHB!

Can you tell us what progress you have made with this problem, and where you need assistance?

I can not prove that it has empty interior! Thank you!
 
  • #4
fsblajinha said:
I can not prove that it has empty interior!
Hint: the complement of $A _ {\epsilon}$ contains every rational point in the interval.
 
  • #5
1 - To show that $A_{\epsilon}$ is measurable, we need to show that it is a subset of a measurable set. Since we are working in the interval $[0,1]$, we know that this interval is a measurable set. By definition, a measurable set is one for which the Lebesgue measure can be defined. Since $A_{\epsilon}$ is a subset of $[0,1]$, it follows that $A_{\epsilon}$ is also a measurable set.

To show that $A_{\epsilon}$ is non-empty, we can choose any rational number $q_{n}$ in the interval $[0,1]$ and construct the interval $I_{n}$ as given in the problem. This interval will always contain a rational number, since we are adding and subtracting a small quantity from $q_{n}$. Therefore, $A_{\epsilon}$ is non-empty.

To show that $A_{\epsilon}$ is empty inside, we need to show that the Lebesgue measure of its interior is zero. Since $A_{\epsilon}$ is a subset of $[0,1]$, its interior is also a subset of $[0,1]$. Since the Lebesgue measure of $[0,1]$ is $1$, it follows that the Lebesgue measure of the interior of $A_{\epsilon}$ is also $1$. Therefore, $A_{\epsilon}$ is empty inside.

2 - To show that $\lambda^{*}(B)=1$, we need to show that the Lebesgue outer measure of $B$ is equal to $1$. By definition, the Lebesgue outer measure of a set $B$ is the infimum of the sums of the lengths of intervals that cover $B$. In this case, we have constructed a sequence of intervals $I_{n}$ that cover $B$. The length of each interval is $\epsilon/2^{n+1}$, and since we are taking the union of infinitely many of these intervals, the total length will approach $1$ as $n$ approaches infinity. Therefore, the Lebesgue outer measure of $B$ is equal to $1$, and we have shown that $\lambda^{*}(B)=1$.
 

FAQ: Is A_epsilon Non-Empty and Measurable in Lebesgue's Theory?

What is the Lebesgue measure in [0,1]?

The Lebesgue measure in [0,1] is a mathematical concept that measures the size or extent of a set of real numbers between 0 and 1. It is a generalization of the more familiar concept of length, and it allows us to measure not just intervals, but any subset of the real numbers.

How is the Lebesgue measure different from other measures?

The Lebesgue measure differs from other measures, such as the Riemann integral, in that it takes into account the "shape" of a set rather than just its size. This allows us to measure more complicated sets, such as fractals, that cannot be measured by traditional methods.

What is the importance of the Lebesgue measure in [0,1]?

The Lebesgue measure is important because it provides a rigorous and flexible framework for measuring sets of real numbers. It has numerous applications in mathematics, physics, and engineering, and it allows us to define and study important concepts such as probability and integration in a more general and powerful way.

How is the Lebesgue measure calculated?

The Lebesgue measure of a set in [0,1] is calculated by dividing the set into smaller and smaller intervals, and then taking the limit of the sum of the lengths of these intervals. This approach is known as the Lebesgue integral, and it is a more general and powerful method than the traditional Riemann integral.

Can the Lebesgue measure be extended to higher dimensions?

Yes, the Lebesgue measure can be extended to higher dimensions. In fact, the Lebesgue measure in [0,1] is just a special case of the Lebesgue measure in higher dimensions, which can be used to measure sets in n-dimensional space. This allows us to generalize the concept of volume to more complex and abstract spaces, and it has numerous applications in fields such as geometry and topology.

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