Due February 9
Problem 1
The Cantor set is totally disconnected (for any $latex x\not= y\in C$ there is $latex z\not\in C$ between x and y) and perfect (compact and without isolated points).
Problem 2
Let $latex E\subset \R$ and
$latex O_n = \{x: d(x,E) < 1/n\}.$
- If E is compact, $latex m(E) = \lim_{n\to\infty} m(O_n)$
- The previous conclusion may be false for E closed and unbounded, or open and bounded.
Problem 3 (Borel-Cantelli lemma)
Let $latex \{E_k\}$ be a sequence of measurable sets such that
$latex \displaystyle \sum_{k=1}^\infty m(E_k) < \infty,$
and define $latex E = \limsup_{k\to\infty} E_k = \{x\in\R^d: x\in E_k$ for infinitely many $latex k\}$.
- E is measurable.
- $latex m(E) = 0$.
Problem 4
Let, for a subset $latex E\subset\R^d$,
$latex \displaystyle m_*^R(E) = \inf \sum_{j=1}^\infty |R_j|,$
where the infimum is taken over all countable covers $latex \{R_j\}$ for E of rectangles. Then
$latex m_*^R(E) = m_*(E).$
We can thus conclude that we obtain the same Lebesgue measure theory if cubes are replaced by rectangles.
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