Proving the existence of a subset, given the Lebesgue outer measure

Question

If c ∈ R, E is a bounded subset of ℝ such that outer Lebesgue measure m*(E) > 0 and 0 < c < m∗(E). How can I show that there exists a subset F of E such that m*(F) = c?

Answer 1

Let $E \subset \mathbb{R}$ be bounded with positive outer measure and let $f: \mathbb{R} \to [0,\infty]$ be defined by $f(x) = m^*((-\infty,x] \cap E)$. Let $x_n$ be a decreasing sequence of real numbers converging to $x$. Then $(-\infty,x] = \bigcap_{n=1}^\infty (-\infty, x_n]$. Using continuity from above, since $m*((-\infty, x_1]\cap E) < \infty$ given that $E$ is bounded, we obtain

\begin{align*}f(x) & = m^*((-\infty,x] \cap E\\ & = m^* \left( \bigcap_{n=1}^\infty (-\infty, x_n]\cap E\right) \\ & = \lim_{n\to \infty} m^*((-\infty, x_n]\cap E)\\ & = \lim_{n\to \infty} f(x_n). \end{align*}

This shows that $f$ is continuous on $\mathbb{R}$. Further, $\lim_{x \to \infty} f(x) = m(E)$ and $\lim_{x\to -\infty} f(x) = 0$. Now, let $0 < c < m^*(E)$. By the Intermediate Value Theorem, choose $x \in \mathbb{R}$ such that $f(x) = c$, that is, $m^*((-\infty,x] \cap E) = c$. Then $F = (-\infty,x] \cap E$ is the desired subset of $E$.

Answer 2

Since $E$ is bounded, it is contained in some interval $[-M, M]$. The desired result will follow if we can show that the function $f(x) = m^*([-M, x] \cap E)$ is continuous, because $f(-M) = m^*(\emptyset) = 0$ and $f(M) = m^*(E) > c$, and thus by the intermediate value theorem there must be some $x_0 \in [-M, M]$ such that $f(x_0) = c$. Then $F = [-M,x_0] \cap E$ has the required properties.

Let $h > 0$. By subadditivity of $m^*$, we have $$\begin{aligned} f(x+h) &= m^*([-M, x+h] \cap E) \\ &\leq m^*([-M, x] \cap E) + m^*((x, x+h] \cap E) \\ &= f(x) + m^*((x, x+h] \cap E) \\ &\leq f(x) + m^*((x, x+h]) \\ &= f(x) + h \end{aligned}$$ and therefore $0 \leq f(x+h) - f(x) \leq h$.

Similarly, if $h < 0$, we have $$\begin{aligned} f(x) &= m^*((-M,x] \cap E) \\ &\leq m^*((-M,x+h] \cap E) + m^*((x+h,x] \cap E) \\ &= f(x+h) + m^*((x+h,x] \cap E) \\ &\leq f(x+h) + m^*((x+h,x]) \\ &= f(x+h) - h \\ \end{aligned}$$ and therefore $h \leq f(x+h) - f(x) \leq 0$.

Combining the above results we conclude that $\lim_{h \to 0}f(x+h) = f(x)$, hence $f$ is continuous as desired.