Assume that
$A = \bigcup_{k=1}^{n_1} I_k$,
$B = \bigcup_{k=1}^{n_2} J_k$,
are disjoint unions with
$I_k, J_k \subset \Bbb R$.
Let $\lambda$ be the Lebesgue-measure. Show that
$\lambda(A) + \lambda(B) \le \lambda(A + B)$.
In order to do so, construct disjoint unions $A = A_1 \cup A_2$ and $B = B_1 \cup B_2$ such that $(A + B) \supset (A_1 + B_1) \cup (A_2 + B_2)$. Then, use induction to show the inequality.
Edit: This whole thing seems to be called "Brunns-Minkowski-inequality".
We already know that
$ a)$ $\lambda(A) + \lambda(B) \le \lambda(A + B)$ with $A = [a_1, b_1), B = [a_2, b_2).$
I didn't solve this either, wrote it down here:
Prove that $\lambda(A + B) \ge \lambda(A) + \lambda(B) $ for A and B being half-open intervals
Now, for this sub-excercise, I started the following way:
For $n = 2$, we get that $n_1 = n_2 = 1$, and thus, we get:
$\lambda(\bigcup_{k=1}^{n_1} I_k) + \lambda(\bigcup_{k=1}^{n_2} J_k) = \lambda(A) + \lambda(B)$, and thus, the inequality follows from $a)$.
For $n \rightarrow n+1$:
We can either work with $n_1 + 1$ or $n_2 + 1$, so let's just define $(n+1) := (n_1 + 1) + n_2$. As long as I understood it so far, we use the same union of $J_k$ for $B$, but a different kind of union of $I_k$ for $A$. Besides that, we are still talking about the same $A$ here. We have to show:
$\lambda(\bigcup_{k=1}^{n_1+1} I_k) + \lambda(\bigcup_{k=1}^{n_2} J_k) \le \lambda(\bigcup_{k=1}^{n_1+1} I_k + \bigcup_{k=1}^{n_2} J_k)$.
Now, my intuitive approach would be to write $\bigcup_{k=1}^{n_1} I_k \cup I_{n+1}$, and with the help of using the induction hypothesis, we would get that this is $\le \lambda(\bigcup_{k=1}^{n_1+1} I_k + \bigcup_{k=1}^{n_2} J_k) + \lambda(I_{n+1})$.
Unfortunately, this is nonsense because we cannot assume that $\lambda(I_{n+1}) = 0$, and thus, we wouldn't get the desired expression. Looking further into it tells me why this is nonsense:
In the induction hypothesis, we assumed that $A = (\bigcup_{k=1}^{n_1} I_k)$. Now, of course we could also assume that $A = (\bigcup_{k=1}^{n_1+1} I_k)$ and say that this is just a different union of $A$, but when we now take a look at $(\bigcup_{k=1}^{n_1} I_k \cup I_{n+1}))$, then it seems like this left union cannot be the same union as the one we assumed to have in the induction hypothesis. This would only hold for $I_{n+1} = \{\emptyset\}$, but I don't think that we are allowed to assume so. So I guess we have to work with these disjoint unions mentioned above, but I don't know how to do it.
I mean, I found that I could construct these disjoint unions with the specific properties by dividing $A$ and $B$ by $2$ (at least I think I did), but I don't see how this helps me here.