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{\bf Question}

In ${\bf R^n}$, show that if $S=\{(x_1, \cdots x_n):x_r=a\}$ then
$m^*(S)=0$.



\vspace{0.25in}

{\bf Answer}

Let $R_n=\{(x_1, \cdots x_n):|x_i|\leq n, \,\,\, i\not=r, \,\,\,
a-\epsilon_n \leq x_r \leq a+\epsilon_n\}$

where $\ds
\epsilon_n=\epsilon\left(\frac{1}{2}\right)^{n+1}\frac{1}{(2n)^{n-1}}$

Then $\ds \bigcup_{n=1}^\infty R_n\supseteq S$

$\ds |R_n|=(2n)^{n-1}2\epsilon_n=\frac{\epsilon}{2^n}
\hspace{0.2in}$ Therefore $\ds \sum_{n=1}^\infty|R_n|=\epsilon$

Thus $m^*(S)=0$



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