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

[fsblajinha]

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 $\displaystyle B:=\bigcup_{k\geq 1}{A_\frac{1}{k}}$

 

[Opalg]

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?

 

[fsblajinha]

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!

 

[Opalg]

I can not prove that it has empty interior!

Hint: the complement of $A _ {\epsilon}$ contains every rational point in the interval.

 

[blue_raver22]

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 $\lambda^{*}(B)=1$.