Prove that $λ^{∗}_{1} (A) + λ^{∗}_{1} (B) ≤ λ^{∗}_{1}(A + B)$

Question

I have this exercise that I don't know if I solved right.

Let $A, B ⊆ \mathbb{R}$ be Borel-measurable subsets with $\max A = 0 = \min B$. We consider $A + B := \{a + b : a ∈ A \text{and} b ∈ B\}$.

Show that $λ^{∗}_{1} (A) + λ^{∗}_{1} (B) ≤ λ^{∗}_{1}(A + B)$.

So what I thought to do is:

I can write $λ^{∗}_{1}(A)+λ^{∗}_{1}(B)\\ =λ^{∗}_{1}(A\cap B^c)+λ^{∗}_{1}(B \cap A^c)+2λ^{∗}_{1}(\{ 0\})\\ = λ^{∗}_{1}(A\backslash \{ 0 \})+λ^{∗}_{1}(B\backslash \{ 0\})+λ^{∗}_{1}(\{ 0\}) + λ^{∗}_{1}(\{ 0\}) $

Since A B and 0 are disjoint we have that $λ^{∗}_{1}(A\backslash \{ 0 \})+λ^{∗}_{1}(B\backslash \{ 0\})+λ^{∗}_{1}(\{ 0\})=λ^{∗}_{1}(A+B)$

So we have $λ^{∗}_{1}(A)+λ^{∗}_{1}(B)=λ^{∗}_{1}(A+B)+λ^{∗}_{1}(\{ 0\})$

But here my results would be $λ^{∗}_{1}(A)+λ^{∗}_{1}(B) \geq λ^{∗}_{1}(A+B)$ which is not what was to prove.

So where are my errors?

Answer 1

I think the idea is to show that there are $a\in A$ and $b\in B$ such that essentially, $a+B\subseteq A+B$ and $A+b\subseteq A+B$ while $(a+B)\cap (A+b)=\emptyset$.

First note that $\lambda^*(A+B)\geq \lambda^*(A)$ and $\lambda^*(A+B)\geq \lambda^*(B)$. Therefore you can assume $\lambda^*(A),\lambda^*(B)<\infty$.

Therefore you can assume that for all $\epsilon>0$ there exists $\alpha,\beta> 0$ such that $$\lambda^*(A) - \lambda^*([-\alpha,0]\cap A)<\epsilon \quad \text{and} \quad \lambda^*(B) - \lambda^*([0,\beta]\cap B)<\epsilon.$$

If $-\alpha \in A$ and $\beta \in B$, then

$$ -\alpha+B^\epsilon\subseteq [-\alpha, \beta-\alpha] \quad \text{and} \quad A^\epsilon+\beta \subseteq [\beta-\alpha, \beta], $$

for $A^\epsilon:=[-\alpha,0]\cap A$ and $B^\epsilon:=[0,\beta]\cap B$, which implies that $(-\alpha+B^\epsilon)\cap (A^\epsilon+\beta)\subseteq \{ \beta -\alpha \}$. So,

$$ \lambda^*(A+B) \geq \lambda^*(-\alpha+B^\epsilon) + \lambda^*(A^\epsilon+\beta)= \lambda^*(A^\epsilon)+ \lambda^*(B^\epsilon). $$

Since $\lambda^*(A\cap [-\alpha,0])$ is monotonic with respect to $\alpha$, you can choose such an $\alpha \in A$. Likewise, $\lambda^*(B\cap [0,\beta])$ is monotonic with respect to $\beta$, so you can choose such a $\beta \in B$.