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{\bf Question}

Define Lebesgue outer measure $m*$ in 3-dimensional Euclidean
space ${\bf R}^3$.  Let $T:{\bf R}^3 \to {\bf R}^3$ be the
transformation defined by $$(x,y,z)\to (hx,ky,lz),$$ where $h,\,k$
and $l$ are nonzero real numbers.  If $E$ is a subset of ${\bf
R}^3$, express $m^*(T(E))$ in terms of $m^*(E)$, proving your
relationship from your dimension of $m^*$.  Show also that if $E$
is measurable then $T(E)$ is measurable.

Find the Lebesgue measure in ${\bf R}^3$ of the ellipsoid
$$\left\{ (x,y,z): \frac{x^2}{a^2}+ \frac{y^2}{b^2} +
\frac{z^2}{c^2}\leq 1\right\}.$$ [The Lebesgue measure of the unit
ball in ${\bf R}^3$ may be assumed.]

\vspace{.25in}

{\bf Answer}

If $E$ is a subset of ${\bf R}^3$, we define Lebesgue outer
measure $m^*$ by $$m^*(E) = \inf_{\{R_i\}} \sum_{i=1}^\infty
|R_i|$$ Where $\{R_i\}$ denotes a system of rectangles of the from
$\{(x_1, x_2, x_3)| a_i &lt; x_1 &lt;b_i \, i = 1, 2, 3\}$ with the
property that $\ds E \subseteq \bigcup_{i=1}^\infty R_i$, and
where $$|R_i| = (b_1 - a_1)(b_2-a_2)(b_3-a_3)$$ Now if $R_i$ is
the rectangle defined by \begin{eqnarray*} a_1 &lt; &amp; x_2 &amp; &lt; b_1
\\ a_2 &lt; &amp; x_2 &amp; &lt; b_2 \\ a_3 &lt; &amp; x_3 &amp; &lt; b_3 \end{eqnarray*} Then
$T(R_i)$ is a rectangle defined by $$A\left\{\begin{array}{rclll}
ha_1 &lt; &amp; x'_2 &amp; &lt; hb_1 &amp; &amp; ({\rm or \ }hb_1 &lt; x_1' &lt; ha_1 {\rm \ \
if \ \ } h &lt;0) \\ ka_2 &lt; &amp; x'_2 &amp; &lt; kb_2 &amp; &amp; ({\rm or \ }kb_1 &lt;
x_1' &lt; ka_1 {\rm \ \ if \ \ } k &lt;0)
\\ la_3 &lt; &amp; x'_3 &amp; &lt; lb_3 &amp; &amp; ({\rm or \ }lb_1 &lt; x_1' &lt; la_1 {\rm \ \
if \ \ } l &lt;0)\end{array}\right.$$  Now if $\epsilon &gt;0$, we can
find a system $\{R_i\}$, such that $$\sum_{i=1}^\infty |R_i| \leq
m^*(E) + \epsilon, $$ by the defination $T(R_i)$ is a rectangle,
and since $\ds \bigcup_{i=1}^\infty R_i \supseteq E$, we have $\ds
\bigcup_{i=1}^\infty T(R_i) \supseteq T(E)$, and so $\ds m^*(T(E))
\leq \sum_{i=1}^\infty |T(R_i)|$

Now by equation $A$, we have \begin{eqnarray*} |T(R_i)| &amp; = &amp;
|h|(b_1 - a_1) |k|(b_2 - a_2)|l|(b_3 - a_3) \\ &amp; = &amp; |hkl| |r_i|
\end{eqnarray*} Hence $\ds m^*(T(E)) \leq |hkl| \sum_{i=1}^\infty
|R_i| \leq |hkl|(m^*(E) + \epsilon)$

Thus \begin{eqnarray} m^*(T(E)) \leq |hkl|m^*(E)\end{eqnarray}

Now consideration of the inverse transformation $T^{-1}$ applied
to the set $T(E)$ shows similarly that

$\ds m^*(E) = M^*(T^{-1}(T(E))) \leq \frac{1}{|hkl|} m^*(T(E))$

i.e. \begin{eqnarray} m^*(T(E)) \geq |hkl|m^*(E) \end{eqnarray}
Thus by (1) and (2), $$m^*(T(E)) = |hkl| m^*(E)$$

${}$

If $E$ is measurable, then for any set $A$, $m^*(A) = M^*(A - E)
+m^*(A \cap E)$

Now let m $B$ be an arbitrary subset of ${\bf R}^3$.  Taking
$A=T^{-1}(B)$ \begin{eqnarray*} m^*(T^{-1}(B)) &amp; = &amp; m^*(T^{-1}(B)
- E) + m^*(T^{-1}(B) \cap E) \\ &amp; = &amp; m^*(T^{-1}(B - T(E))) +
m^*(T^{-1} (A\cap T(E))) \end{eqnarray*}  Thus we have $$
\frac{1}{|hkl|} m^*(B) = \frac{1}{|hkl|} m^*(B - T(E)) +
\frac{1}{|hkl|} m^*(B \cap T(E))$$ Hence, cancelling $\ds
\frac{1}{|hkl|}$, $T(E)$ is measurable.

Let $\ds E = \left\{(x,y,z)| \frac{x^2}{a^2}+ \frac{y^2}{b^2} +
\frac{z^2}{c^2} \leq 1\right\}$

Let $\ds S = \{ (x,y,z)\, |\,  x^2 + y^2 +z^2 \leq 1 \}$

Let $\ds T : (x,y,z) \rightarrow (|a|x, |b|y, |c|z)$

Then $T(S) = E$

Hence $m^*(E) = |abc|m^*(S)$

$S$ and $E$ are closed and so measurable.  $S$ so the unit sphere,
so $m(S) = \frac{4}{3}\pi$

Hence $m(E) = \frac{4}{3} \pi |abc|$



\end{document}
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