# Show that there is a Lebesgue measurable subset with measure $a$

If $A\subset \mathbb{R}^d$ has positive Lebesgue measure, then for each $0\leq a<m(A)$ there is a Lebesgue measurable $E\subset A$ with $m(E)=a$.

We have that

$E\subset A \Rightarrow m(E)\leq m(A)$

and

$m(A)>a$

How can we conclude that $m(E)=a$ ??

Define $f(t)=m(B_t(0)\cap A)$ where $B_t(0)$ is the open ball of radius $t$ centered at the origin. Show that $f$ is continuous and use the intermediate value theorem.
Define $E_t=A\cap B_t(0)$. Then the measure of $E_t$ is $f(t)$. $f(0)=0$ and $f(t)$ approaches $m(A)$ as $t$ approaches infinity, so for any $a<m(A)$ there exists a $t$ such that $m(E_t)=a$. $E_t$ therefore satisfies the condition.
To show continuity, choose $t_0>0$. We know that the volume of a ball is a continuous increasing function of the radius. Suppose the volume of a ball of radius $t$ is $V(t)$. We are given $\epsilon$ and we want to find $\delta$ such that if $|t-t_0|<\delta$ then $|f(t)-f(t_0)|<\epsilon$. By additivity of the measure, we have that if $p>q$ then $f(p)-f(q)=|f(p)-f(q)|=m(E_p-E_q)=m(A\cap (B_p(0)-B_q(0)))\leq V(p)-V(q)$, with the last inequality given by monotonicity. Choose $\delta$ so that if $|t-t_0|<\delta$, then $|V(t)-V(t_0)|<\epsilon$. Then $|f(t)-f(t_0)|<\epsilon$, so we are done.
• @MattS To apply the intermediate value theorem we do the following: $$0\leq a<m(A)\Rightarrow \exists E\subset A \text{ such that } m(E)=a$$ Is this correct?? At the inequality $0\leq a<m(A)$ the $0$ is the value of measure right?? – Mary Star Nov 10 '14 at 14:41
• Perhaps using $[-t,t]^d$ instead of $B_t(0)$ would make it easier to show continuity. – Henning Makholm Nov 10 '14 at 14:53