If $m^*([-n,n] \cap E) + m^*([-n,n] \setminus E) = 2n$ for all $n$, then $E$ is Lebesgue measurable

Let $E \subset \Bbb R$ and let $m^*$ denote the Lebesgue outer measure on $\Bbb R$. Show that if for all $n \in \Bbb N$, $m^*([-n,n] \cap E) + m^*([-n,n] \setminus E) = 2n$, then $E$ is Lebesgue measurable.

Hint: for all $n \in \Bbb N$, there are measurable sets $F_n$, $G_n$ such that $F_n \subset E \cap [-n,n] \subset G_n$, and $m(G_n \setminus F_n) = 0$.

Given the hint, it is not difficult to conclude: put $F = \cup_n F_n$ and $G = \cup_n G_n$. We have:

$$m^*(E \setminus F) = m^* \left( \bigcup_n ((E \cap [-n,n]) \setminus F) \right) \le m (G \setminus F) \le m(\cup_n (G_n \setminus F_n)) = 0$$

So $E \setminus F$ is negligible, hence measurable. Therefore, $E = (E \setminus F) \cup F$ is measurable.

How to prove the claim in the hint?

Solution 1:

To improve readability, we will first fixed our space to $X:=[-n,n]$, any subset of $X$ that is measurable is also measurable in $\Bbb R$. For any $A\subset X$, $A^c$ denotes its complement in $X$. Any arbitrary $E$ satisfying $$ m^*(E)+m^*(E^c)=m^*(X), $$ we'll show that there exist $F,G\subset \Bbb R$ such that $F\subset E \subset G$ and $m(G\backslash F)=0$.

Recall that $$ m^*(E)=\inf\{\ \sum_{i=1}^{\infty}m(A_i) \ |\ E\subset \bigcup_{i=1}^{\infty}A_i\ ;\ A_i\ \text{is open in $X$}\ \}. $$

So for any $\varepsilon>0$, there is a sequence $(A_i^{\varepsilon})$ such that $E\subset \bigcup_{i=1}^{\infty}A_i^{\varepsilon}$ and that $m^*(E)\le \sum_{i=1}^{\infty}m(A_i^{\varepsilon})<m^*(E)+\varepsilon$.

Let $\varepsilon=\frac 1n$ (this $n$ is arbitrary, not the fixed integer used to define $X$), defined $$ G_n:=\bigcup_{i=1}^{\infty}A_i^{1/n}. $$ Observe that $m^*(G_n)\le \sum_{i=1}^{\infty}m(A_i^{1/n})<m^*(E)+\frac 1n$.

Let $$G:=\bigcap G_n,$$ since $E\subset G_n$, $E\subset G$. Therefore $m^*(E)\le m^*(G) \le m^*(G_n)<m^*(E)+\frac 1n$ for all $n\in\Bbb N$, so $$ m^*(G)=m^*(E). $$ Since $G$ is Borel (in $\Bbb R$), it is measurable (in $\Bbb R$).

Similarly, we can do the same for $E^c$ . Let $H_n$ be an open covering of $E^c$ such that $m^*(E^c)\le m^*(H_n) < m^*(E^c)+\frac 1n$. Define $$ H:=\bigcap H_n, $$ we have $m^*(H)=m^*(E^c)$, also. Since $E^c\subset H$, $H^c\subset E$.

Define $F:=H^c$, it is not hard to see that $F$ is Borel (in $\Bbb R$), hence measurable (in $\Bbb R$). This implies that $$ m^*(F)+m^*(F^c)=m^*(X) $$ so we have $$ m^*(F)=m^*(X)-m^*(F^c)=m^*(X)-m^*(H)=m^*(X)-m^*(E^c)=m^*(E). $$

Finally, since $F\subset E\subset G$, we have $$\begin{align} m^*(G\backslash F) &= m^*(G) - m^*(F) \\ &=m^*(E)-m^*(E) \\ &=0. \end{align}$$