Disjoint subsets $A$ and $B$ of $\mathbb{R}^n$ such that $\lambda^*(A\cup B)<\lambda^*(A)+\lambda^*(B)$

Question

Prove that there exist disjoint subsets $A$ and $B$ of $\mathbb{R}^n$ such that $$\lambda^*(A\cup B)<\lambda^*(A)+\lambda^*(B)$$ and $$\lambda_*(A\cup B)>\lambda_*(A)+\lambda_*(B).$$

This is an exercise from Jones' Lebesgue Integration in Euclidean Space. It seems like I should invoke the Axiom of Choice or something like that, but I just can't see how I should proceed.

Above, $\lambda^*$ denotes the outer measure(outer approximation by open sets) and $\lambda_*$ denotes the inner measure(inner approximation by compact sets). $\lambda$ is the lebesgue measure.

Answer 1

Take $A$ any non-measurable subset of $[0,1]$ and $B$ its complement. If we had equality for the outer measures then $A$ would be measurable ( not a tautology, it requires some little argument).

Now use $\mu_{*}(A) = 1 - \mu^*(B)$, and get the required inequality for the inner measures

Answer 2

To prove the first property, we will use for convenience rectangles as sets defined as follows:

An n-dimensional, closed rectangle with sides oriented parallel to the coordinate axes, or rectangle for short, is a subset $R\subset {\mathbb{R}}^n $ of the form, $R=\left[a_1,b_1\right]\times \left[a_2,b_2\right]\times \dots \times \left[a_n,b_n\right]$ where, $-\infty <a_i\le b_i<\infty$.

We denote that the collection of all n-dimensional rectangles by $\mathcal{R}({\mathbb{R}}^n)$ and then $R\to \lambda (R)$ defines a map:

$\lambda :\mathcal{R}\left({\mathbb{R}}^n\right)\to [0,\infty )$, where $\lambda $ is the Lebesgue measure, and the Lebesgue outer measure of a subset$E\subset {\mathbb{R}}^n$ is given from the follow equation: ${\lambda }^*\left(E\right)={\mathrm{inf} \left\{\sum^{\infty }_{i=1}{\lambda \left(R_i\right)}:E\subset \bigcup^{\infty }_{i=1}{R_i},\ \ \ R_i\in \mathcal{R}\left({\mathbb{R}}^n\right)\right\}\ }$.

The Lebesgue outer measure has the following properties:

1. ${\lambda }^*\left(\emptyset \right)=0$
2. If $E\subset F$, then ${\lambda }^*(E)\le {\lambda }^*(F)$
3. If $\left\{E_i\subset {\mathbb{R}}^n:i\in \mathbb{N}\right\}$ is a countable collection of subsets of ${\mathbb{R}}^n$ then, ${\lambda }^*\left(\bigcup^{\infty }_{i=1}{E_i}\right)\le \sum^{\infty }_{i=1}{{\lambda }^*(E_i)}$

The proof of the third property is the proof you need in order to prove your first with the Lebesgue outer measure. So, we have:

If ${\lambda }^*\left(E_i\right)=\infty $ for some $i\in \mathbb{N}$, then there is nothing to prove. So, we assume that ${\lambda }^*(E_i)$ is finite for every $i\in \mathbb{N}$. If $\varepsilon >0$, there is a countable covering $\{R_{ij}:j\in \mathbb{N}\}$ of $E_i$ by rectangles $R_{ij}$ such that

$\sum^{\infty }_{j=1}{\lambda (R_{ij})}\le {\lambda }^*\left(E_i\right)+\frac{\varepsilon }{2^i}$, and $E_i\subset \bigcup^{\infty }_{j=1}{R_{ij}}$. Then $\{R_{ij}:i,j\in \mathbb{N}\}$ is a countable covering of $E=\bigcup^{\infty }_{i=1}{E_i}$.

Therefore,

${\lambda }^*\left(E\right)\le \sum^{\infty }_{i,j=1}{\lambda \left(R_{ij}\right)}\le \sum^{\infty }_{i=1}{\left\{{\lambda }^*\left(E_i\right)+\frac{\varepsilon }{2^i}\right\}}=\sum^{\infty }_{i=1}{{\lambda }^*\left(E_i\right)}+\varepsilon$

Since, $\varepsilon >0$ is arbitrary, it follows that ${\lambda }^*\left(E\right)\le \sum^{\infty }_{i=1}{{\lambda }^*\left(E_i\right)}$.

Note: For $i=1,2$ you have $E=E_1\cup E_2$. By setting $E_1=A$ and $E_2=B$ you have the result you want.