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{\bf Question}

Prove that the Lebesgue measure of the interval $[a,b]$ is $(b-a)$
[in {\bf R}].



\vspace{0.25in}

{\bf Answer}

We shall use open covers by intervals to generate $m^*$.  Let
$\{R_i\}$ be an open cover of $[a,b]$.  There is a finite subcover
$\{R_1,\cdots R_n\}$ which we may suppose ordered so that
$R_i=[a_i,b_i]$ and $a_1\leq a_2\leq \cdots \leq a_n$, also
$a_1<a$ and $a_n>b$, and so that no interval is entirely contained
within another, then $b_i\leq b_j$ for $i<j$, for otherwise
$(a_i,b_i)\supseteq(a_j,b_j)$

also $b_i>a_{i+1}$ for otherwise the point $\ds
\frac{b_i+a_{i+1}}{2}$ is not covered by the intervals.

Therefore $\sum|R_i|=\sum(b_i-a_i)$

$=b_n(-a_n+b_{n-1})(-a_{n-1}+\cdots)+b_1-a_1$

$\geq b_n-a_1 \geq b-a$

Therefore $m^*([a,b])\geq b-a \hspace{0.5in}$  Cover $[a,b]$ by
$[a,b]$

Therefore $m^*([a,b])=b-a$





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