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{\bf Question}

Let $E$ be a bounded measurable subset of the plane.  Let $H(t)$
be the half-plane defined by

$$H(t)=\{(x,y)|x\leq t\}.$$

The function $f(t)$ is defined for all real $t$ by

$$f(t)=m(E\cap H(t)),$$

where $m$ denotes Lebesgue measure in the plane.

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Prove that $f$ is a continuous function, and deduce that there is
a line in the plane parallel to the y-axis which bisects $E$ in
the sense that a subset of $E$ of measure $\frac{1}{2}m(E)$ lies
on each side of the line.

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Give an example of a set $E$ for which the set
$f^{-1}(\frac{1}{2}m(E))$ contains more than one point.

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Show that for all sets $E$, either $f^{-1}(\frac{1}{2}m(E))$ is a
single point or it is a closed interval.


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{\bf Answer}

$E$ is bounded and so $E$ is a subset of some square S with sides
of length $l$ parallel to the co-ordinate axes.

If $t_1<t_2$ then $E\cap H(t_1)\subseteq E\cap H(t_2)$ and so
$f(t_1)\leq f(t_2)$

Then $f(t_2)-f(t_1)=m(E\cap H(t_2))-m(E\cap H(t_1))$

$=m[(E\cap H(t_2))-(E\cap H(t_1))]=m[E\cap(H(t_2)-H(t_1)]$

$\leq m(S\cap(H(t_2)-H(t_1)))=l(t_2-t_1).$

Thus $f$ is continuous everywhere (in fact $f$ is uniformly
continuous).

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Since $E\subseteq S$, if $t$ is large and negative $f(t)=0$ and if
$t$ is large and positive $f(t)=m(E)$.  Thus, by the intermediate
value theorem for continuous functions there is a number $t_0$
such that

$$f(t_0)=\frac{1}{2}m(E)$$

so to the left of the line $x=t_0$ lies a portion of $E$ of
measure $\frac{1}{2}m(E)$ (since the measure of $E\cap l$ is zero
for any line $l$) and also to the right of this line lies a
portion of $E$ of measure $\frac{1}{2}m(E)$.

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Example.  If $E_1$ is the unit disc centre $-2$ and $E_2$ is the
unit disc centre $+2$, and $E=E_1\cup E_2$ then
$f(-1)=f(+1)=\frac{1}{2}m(E)$.

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Now suppose $f^{-1}(\frac{1}{2}m(E))$ is not a single point.

Let $a=\inf\{x|x\epsilon f^{-1}(\frac{1}{2}m(E))\}\,\,\,\,
b=\sup\{x|x\epsilon f^{-1}(\frac{1}{2}m(E))\}$, then $a<b$.  We
can find a sequence $x_n\rightarrow a$ such that $x_n\epsilon
f^{-1}(\frac{1}{2}m(E))$ i.e. $f(x_n)=\frac{1}{2}m(E)$.  Since $f$
is continuous $f(a)=\lim f(x_n)=\frac{1}{2}m(E)$  i.e. $a\epsilon
f^{-1}(\frac{1}{2}m(E))$. Similarly $b\epsilon
f^{-1}(\frac{1}{2}m(E))$.  Since $f$ is an increasing function.

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$f(x)=\frac{1}{2}$ for all $x$ satisfying $a\leq x\leq b$ and so
$f^{-1}(\frac{1}{2})=[a,b]]$.




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