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## Due February 9

### Problem 1

The Cantor set is totally disconnected (for any $x\not= y\in C$ there is $z\not\in C$ between x and y) and perfect (compact and without isolated points).

### Problem 2

Let $E\subset \R$ and

$O_n = \{x: d(x,E) < 1/n\}.$

1. If E is compact, $m(E) = \lim_{n\to\infty} m(O_n)$

2. The previous conclusion may be false for closed and unbounded, or open and bounded.

### Problem 3 (Borel-Cantelli lemma)

Let $\{E_k\}$ be a sequence of measurable sets such that

$\displaystyle \sum_{k=1}^\infty m(E_k) < \infty,$

and define $E = \limsup_{k\to\infty} E_k = \{x\in\R^d: x\in E_k$ for infinitely many $k\}$.

1. E is measurable.

2. $m(E) = 0$.

### Problem 4

Let, for a subset $E\subset\R^d$,

$\displaystyle m_*^R(E) = \inf \sum_{j=1}^\infty |R_j|,$

where the infimum is taken over all countable covers $\{R_j\}$ for E of rectangles. Then

$m_*^R(E) = m_*(E).$

We can thus conclude that we obtain the same Lebesgue measure theory if cubes are replaced by rectangles.