Lebesgue Measure

Problem 1

Let \(A_n\) be any sequence of subsets of \(\mathbb{R}\)
(i) if \(A_n\uparrow A\) then \(\lambda^*(A_n)\uparrow \lambda^*(A)\)
(ii)\(\lambda^*(\liminf A_n)\leq\liminf\limits_{n\to\infty} \lambda^*(A_n)\), where \(\liminf A_n=\{x:x\in A_n \mbox{ultimately}\}\)

solution:
(i)
Let  \(F_n=\bigcup_{i=1}^{n}E_i\), \(F=\cup E_n,A=\cup A_n\), where \(E_n\) is measurable such that \(A_n\subset E_n\), and \(E_n-A_n\) is negligible (it is easy to prove such \(E_n\) exists), clearly \(A\subset F\),
\(F_n\uparrow F\), therefore \(\lambda^*(F_n)\uparrow \lambda^*(F)\)
\(\begin{align}
\lambda^*(A_n)\leq\lambda^*(F_n)&\leq\lambda^*(A_n)+\lambda^*(F_n-A_n)\\
&\leq\lambda^*(A_n)+\sum_{i=1}^{n}\lambda^*(E_i-A_i)\\
&= \lambda^*(A_n)
\end{align}\)
\(\lambda^*(F_n)=\lambda^*(A_n)\leq \lambda^*(A)\leq \lambda^*(F)\), and since \(F_n\uparrow F\), we have \(\lambda^*(A_n)\uparrow \lambda^*(A)\)
(ii)
Let \(B_n=\bigcap_{k\geq n}A_k\), then \(B_n\uparrow B=\liminf A_n\)
\(\lambda^*(B_n)\leq \lambda^*(A_n)\Rightarrow \liminf\limits_{n\to\infty} \lambda^*(B_n)=\lambda^*(\liminf A_n)\leq \liminf\limits_{n\to\infty} \lambda^*(A_n)\)
The equality follows from (i)

Problem 2

For \(A\subset [a,b]\subset \mathbb{R}\), \(\lambda^*(A)\) is also called the exterior measure of A in \([a,b]\) and is denoted \(\lambda_e(A)\), the interior measure of A in \([a,b]\), denoted \(\lambda_i(A)\), is defined by the formula
\[\lambda_i(A)=(b-a)-\lambda_e([a,b]-A)\]in general, \(\lambda_i(A)\leq\lambda_e(A)\) ; the set A is measurable if and only if \(\lambda_i(A)=\lambda_e(A)\)

solution:
\(\Rightarrow\) is easy
\(\Leftarrow\) We use \(\lambda^*\) instead of \(\lambda_e\) denote \(I=[a,b]\), let \(J\) be any bounded interval \begin{align*} \lambda^*(I)&=\lambda^*(I\cap A)+\lambda^*(I\cap A')\\ &=\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A\cap J')+\lambda^*(I\cap A'\cap J)+\lambda^*(I\cap A'\cap J') \\
&=\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A'\cap J)+\lambda^*(I\cap A\cap J')+\lambda^*(I\cap A'\cap J')\\
&\geq \lambda^*(I\cap J)+\lambda^*(I\cap J')\\
&=\lambda^*(I)
\end{align*}
Therefore we have \(\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A'\cap J)=\lambda^*(I\cap J)\), and it is easy to see that \(\lambda^*(A\cap J)+\lambda^*(A'\cap J)=\lambda^*(J)\) Suppose \(S\) is a subset of \(\mathbb{R}\) for every \(\varepsilon>0\), there exists \(U,S\subset U=\bigcup J_n\) and \(\sum \lambda^*(J_n)<\lambda^*(S)+\varepsilon\)
\begin{align*}
\lambda^*(A\cap S)+\lambda^*(A'\cap S)&\leq\lambda^*(A\cap U)+\lambda^*(A'\cap U)\\
&\leq \sum\lambda^*(A\cap J_n)+\sum\lambda^*(A'\cap J_n)\\ &=\sum(\lambda^*(A\cap J_n)+\lambda^*(A'\cap J_n))\\
&=\sum\lambda^*(J_n)\\
&<\lambda^*(S)+\varepsilon
\end{align*}
Since \(\varepsilon\) is arbitrary, we can conclude that A is measurable 

Problem 3

If A is a Lebesgue-measurable subset of \(\mathbb{R}\) such that \(\lambda(A)>0\), then the difference set \(D=\{a-b:a,c\in A\}\) is a neighborhood of 0, that is, \((-\delta,\delta)\subset D\) for some \(\delta>0\).