Prove that $\lambda(A + B) \ge \lambda(A) + \lambda(B) $ for A and B being half-open intervals 
Prove that for the Lebesgue-measure $\lambda$, the inequality
$\lambda(A + B) \ge \lambda(A) + \lambda(B) $
holds.

So, this task is divided in some smaller tasks, and I'm supposed to begin with the following one:

Show that the inequality holds for half-open intervals
$A = [a_1, b_1)$ and $B = [a_2, b_2)$
with $a_1, a_2, b_1, b_2 \in \Bbb R$, $b_1 > a_1$ and $b_2 > a_2$.

I tried to do a case analysis, taking a look at $A \subset B$, $A \cap B = \emptyset$ and $A \cap B \neq \emptyset$. Yet, I only came to the conclusion that, no matter what case I try to prove, that there is always only the equality between $A$ and $B$.
So, we know that $A + B = [a_1, b_1) + [a_2, b_2) = [a_1 + a_2, b_1 + b_2)$. This leads to
$\lambda(A + B) = \lambda([a_1 + a_2, b_1 + b_2)) = (b_1 + b_2) - (a_2 + a_1) = b_1 + b_2 - a_2 - a_1 = b_1 - a_1 + b_2 - a_2 = \lambda([a_1, b_1)) + \lambda([a_2, b_2)) = \lambda(A) + \lambda(B).$
Of course this doesn't contradict the statement above, but I guess there is more to show here. We are allowed to use the usual properties of a measure.
Edit:
So, basically, I couldn't find an example where the equality didn't hold. I guess this is exactly sub-task wats me to find? We later work with $A$ and $B $ being compact, so I might get a different result, but I'm not sure.
EditEdit:
It turned out that my solution is perfectly fine, I simply refused to believe that the equation would hold for every interval - but it does.
 A: For the strick inequality you may take $A=B=C$ where C is the standard Cantor set. We know $C+C=[0,2]$ and $\lambda(C)=0$ so here $$\lambda(C+C) > \lambda(C)+\lambda(C)$$
Or, Just take $A=(0,1) \cup (2,3)$ and $B=(1,2)$ then $A+B=(1,5)$ here $\lambda(A+B)=4$ but $\lambda(A)+\lambda(B)=3$.
A: Assuming that $A\cap B=\emptyset $. Then since $A,B$ are compact so $d(A,B)=\inf \{|a-b|:a\in A,b\in B\}>0$ say $d(A,B)=r>0$
Let $\epsilon >0$ .Then $\exists $ sequence $\{I_n\}$ disjoint open intervals each with diameter $<r$ such that $A+B\subseteq \cup I_n$ and $\lambda(A+B)+\epsilon>\sum l(I_n)$
Now $A\subseteq A+B;B\subseteq A+B\implies A\subseteq \cup I_n,B\subseteq \cup I_n$.
Let $\{I_n\}^{'}$ denote those intervals which contain only points of $A$,and Let $\{I_n\}^{''}$ denote those intervals which contain only points of $B$.
$A\subseteq \cup I_n^{'},B\subseteq \cup I_n^{''}$.
Then $\lambda (A)+\lambda (B)\le \sum l(I_n^{'})+\sum l(I_n^{''})\le \sum I_n<\lambda(A+B)+\epsilon$
which holds $\forall \epsilon >0$ .Hence the result
A: It is not much to harvest here on the line. For the Lebesgue measure on $\Bbb{R}$ it is always equality. However, even on $\Bbb{R}^2$ you can simply get strict inequality, for example, for $A=B=[0,1)^2$ $\Rightarrow$ $A+B=[0,2)^2$ and
$$
\lambda(A+B)=4>1+1=\lambda(A)+\lambda(B).
$$
