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 Post subject: On Measure Theory
PostPosted: Fri Jul 08, 2016 1:16 pm 
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(1) Let \( \displaystyle \left( X , \mathcal{A} , \mu \right) \) be a complete measure space. If for some \( \displaystyle A \in \mathcal{A} \) and some \( \displaystyle B \subset X \) holds \( \displaystyle A \bigtriangleup B \in \mathcal{A} \text{ and } \mu \left( A \bigtriangleup B \right) = 0 , \) show that \( \displaystyle B \in \mathcal{A} \text{ and } \mu \left(B\right) = \mu \left(A\right). \)

(2) Let \( \displaystyle Q= \left\{ q_{n} | n\in\mathbb{N} \right\} \) be an enumeration of \( \displaystyle \mathbb{Q} \cap \left[0,1\right] \).
For every \( \displaystyle \epsilon > 0 \) define
\[ \displaystyle A_{\epsilon} = \bigcup_{n=1}^{\infty} \left( q_{n} - \frac{\epsilon}{2^{n+1}} \, , \, q_{n} + \frac{\epsilon}{2^{n+1}} \right) \]
Also, define \[ \displaystyle A = \bigcap_{j=1}^{\infty} A_{\frac{1}{j}} \]
Let \( \displaystyle \lambda \) be the Lebesgue measure on \( \mathbb{R}. \)

(i) Show that \( \displaystyle \lambda \left( A_{\epsilon} \right) \leq \epsilon. \)

(ii) If \( \epsilon < 1 \) show that the set \( \displaystyle \left[0,1\right] \setminus A_{\epsilon} \) is nonempty.

(iii) Show that \( \displaystyle A \subset \left[0,1\right] \, , \, Q \subset A \, \text{ and } \lambda(A)=0. \)


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 Post subject: Re: On Measure Theory
PostPosted: Fri Jul 08, 2016 1:18 pm 
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(1) Recall that
\[ \displaystyle A \bigtriangleup B = \left(A \smallsetminus B \right) \cup \left( B \smallsetminus A \right) \]
\( \bullet \) Since
\[ \displaystyle A \bigtriangleup B \in \mathcal{A}, \; A \smallsetminus B , B \smallsetminus A \subset A \bigtriangleup B , \; \mu \left( A \bigtriangleup B \right) = 0 \]
the sets \( \displaystyle A \smallsetminus B , B \smallsetminus A \) are \( \mu \)-null. Thus,
\[ \displaystyle A \smallsetminus B , B \smallsetminus A \in \mathcal{A}, \]
because the measure space has been supposed being complete.

\( \bullet \) Since
\[ \displaystyle A \cup B = A \cup \left( B \smallsetminus A \right), \; A \in\mathcal{A}, \; B \smallsetminus A \in \mathcal{A} \]
and \( \mathcal{A} \) is a \( \sigma - \)algebra on X, \[ \displaystyle A \cup B \in \mathcal{A}. \]
\( \bullet \) Since
\[ \displaystyle A \cap B = \left( A \cup B \right) \smallsetminus \left( A \bigtriangleup B \right), \; A \cup B \in \mathcal{A}, \; A \bigtriangleup B \in \mathcal{A} \] and \( \mathcal{A} \) is a \( \sigma - \)algebra on X, \[ \displaystyle A \cap B \in \mathcal{A}. \]
Now, notice that \( \displaystyle B = \left( B \smallsetminus A \right) \cup \left( A \cap B \right), \) and since these two sets belong to \( \mathcal{A} \) and \( \mathcal{A} \) is a \( \sigma - \)algebra on X, \( \displaystyle B \) belongs to \( \mathcal{A} \), too. Therefore we have proved the first assertion.


Note that
\[ \displaystyle A = \left( A \smallsetminus B \right) \cup \left( A \cap B \right) \]
\[ \displaystyle B = \left( B \smallsetminus A \right) \cup \left( A \cap B \right) \]
\[ \displaystyle A \bigtriangleup B = \left(A \smallsetminus B \right) \cup \left( B \smallsetminus A \right) \]
and the above unions consist of disjoint sets. Since \( \mu \) is a measure on X, it follows that
\[ \displaystyle \mu \left(A\right) = \mu \left( A \smallsetminus B \right) + \mu \left( A \cap B \right) \]
\[ \displaystyle \mu \left(B\right) = \mu \left( B \smallsetminus A \right) + \mu \left( A \cap B \right) \]
\[ \displaystyle \mu \left( A \bigtriangleup B\right) = \mu \left( A \smallsetminus B \right) + \mu \left( B \smallsetminus A \right), \]
respectively. Since it has been supposed that \( \displaystyle \mu \left( A \bigtriangleup B \right) = 0 \), we have that \( \displaystyle \mu \left( A \smallsetminus B \right) = \mu \left( B \smallsetminus A \right) = 0 \). Hence,
\[ \displaystyle \mu \left(A\right) = \mu \left(B\right) \]
and the second assertion has been proved as well.


P.S. The second part of the exercises, that is (2), has not yet been answered. I have faith that someone will give a nice solution to this tricky exercise!


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 Post subject: Re: On Measure Theory
PostPosted: Sun Jul 17, 2016 9:52 pm 

Joined: Thu Jan 28, 2016 8:52 pm
Posts: 9
Hello everyone. For exercise (2):

(i) We have that $\lambda (A_\varepsilon )=\lambda \left (\bigcup_{n=1}^{\infty}\left ( q_n-\frac{\varepsilon }{2^{n+1}},q_n+\frac{\varepsilon }{2^{n+1}} \right ) \right )\leq \sum_{n=1}^{\infty}\lambda \left ( q_n-\frac{\varepsilon }{2^{n+1}},q_n+\frac{\varepsilon }{2^{n+1}} \right ) = \sum_{n=1}^{\infty} \frac{\varepsilon }{2^n}=\varepsilon $.


(ii) Suppose that $\varepsilon <1$ and $\left [ 0,1 \right ]\setminus A_\varepsilon = \varnothing $. Then $\left [ 0,1 \right ]\subseteq A_\varepsilon \Rightarrow 1=\lambda (\left [ 0,1 \right ])\leq \lambda (A_\varepsilon )\leq \varepsilon $, a contradiction.


(iii) It suffices to show that every $A_{\frac{1}{j}}$ is contained in $\left [ 0,1 \right ]$, which is easy to see. Since $0\leq q_n\leq 1$, for every $j\geq 1$, we have $x \in A_{\frac{1}{j}}=\left (q_n-\frac{\frac{1}{j}}{2^{n+1}},q_n+\frac{\frac{1}{j}}{2^{n+1}} \right )\Rightarrow -\frac{1}{j}\leq q_n-\frac{\frac{1}{j}}{2^{n+1}}\leq x\leq q_n+\frac{\frac{1}{j}}{2^{n+1}}\leq \frac{1}{j}$. So $ A_{\frac{1}{j}}\subseteq \left [ 0,1 \right ]$, for every $j\geq 1\Rightarrow A=\bigcap_{j=1}^{\infty}A_{\frac{1}{j}}\subseteq \left [ 0,1 \right ]$.

Also, $Q=\mathbb{Q}\cap \left [ 0,1 \right ]\subseteq A_\frac{1}{j}$, since $q_n\epsilon A_\frac{1}{j}$ for every $n,j\geq 1$. Thus $Q\subseteq \bigcap_{j=1}^{\infty}A_{\frac{1}{j}}=A$.

Lastly, using (i) we get that $\lambda (A)\leq \lambda \left (A_{\frac{1}{j}} \right )\leq \frac{1}{j}$ and that is true for every $j\geq 1$. Hence, $\lambda (A)=0$.


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 Post subject: Re: On Measure Theory
PostPosted: Sun Jul 17, 2016 9:57 pm 

Joined: Thu Jan 28, 2016 8:52 pm
Posts: 9
We can also see that $A$ is uncountable.

Suppose that $A$ is countable and $A=\left \{ x_n:n\epsilon \mathbb{N} \right \}$, then we write $\left [ 0,1 \right ]=A\cup \left (\left [ 0,1 \right ]\setminus A \right )=\left (\bigcup_{n=1}^{\infty}\left \{ x_n \right \} \right )\cup \left (\bigcup_{j=1}^{\infty}\left (\left [ 0,1 \right ]\setminus A_\frac{1}{j} \right ) \right )$. Since $\left \{ x_n \right \}$ and $\left [ 0,1 \right ]\setminus A_\frac{1}{j}$ are closed for every $n,j\geq 1$, Baire's theorem guarantees that one of these sets must have non-empty interior. The set $\left \{ x_n \right \}$ clearly cannot contain any interval for any $n\geq 1$. The same happens with $ \left [ 0,1 \right ]\setminus A_\frac{1}{j}$, for any $j\geq 1$, since it doesn't contain any rational numbers.

So we reached a contradiction. Thus, $A$ is uncountable.


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