Is this statement implying that a binary variable can be between 0 and 1? I am difficulty understanding the use of a binary decision variable in this paper on page 9: http://www.eecs.qmul.ac.uk/~khouzani/Papers/ESORICS16Techrep.pdf
In section 4 at the bottom of the page in equation (10), it is saying 

“Common to all of our models is the introduction of binary decision
  variables as follows: $x_{cl} \in {0,1}$ for each $c \in C$ and $l \in
> L_c$, which represents whether control c is implemented at level $l
> \in L_c$.  Using this notation, we first enforce that logically at
  most only one of the implementation levels per each control is
  selected: 
$(x_{cl} \in \{0,1\} \forall l \in L_c, \forall c \in C)$
       $\,\,\, \,\,\, \sum\limits_{l \in L_c}x_{cl}\le1, \forall c\in C$
Recall that $L_c := {1,…,L_c}$, and in particular, it did not include
  level 0."

There are a few things I do not understand about this:
1)  Since the decision variable $x_{cl}$ is binary, I do not understand the need for the inequality.  Since it says "at most only one of the implementation levels per each control is selected", I don't understand why we cannot just use this: 
$x_{cl}\in\{0,1\} \,\,\, \forall c\in C$.  Why isn't that used?
2)  Secondly, I am not understanding how the following statement affects this problem formulation: 
"Recall that $L_c := {1,…,L_c}$, and in particular, it did not include level 0."  Provided the decision variable is either 0 or 1, with 0 meaning it is not implemented and 1 meaning it is, then I do not see how this makes any difference?  Does it?
 A: I happen to be one of the authors of that paper (small world indeed!).
1) As it is already correctly pointed out by @Thoth, this is to ensure that at most one of the levels are selected (so either none of the levels, or exactly one of the levels). In particular, it is not enough that each of them  is binary, because without this constraint, it would be permissible that more than one of the levels be picked. For instance, we can have $x_{c1}=1$ or $x_{c2}=1$, but not $x_{c1}=1$ and $x_{c2}=1$ at the same time (although $x_{c1}=0$ and $x_{c2}=0$ is allowed, so that if the optimisation decides not to select a control at all, it can!).
2) Here is the idea: the fact that $\mathcal{L}_c=\{1,\ldots,L_c\}$ allows the option of "no levels of a control c being selected" as $x_{cl}=0$ for all $l\in\mathcal{L}_c$, which satisfies the inequality constraint of $$\sum_{l\in\mathcal{L}_c}x_{cl}\leq 1,\quad \forall c\in\mathcal{C}$$ since the left-hand-side is zero. Now, consider an alternative modelling in which we would explicitly included a level 0 for each control, where the interpretation of level zero for a control is not selecting that control at all. That is, we could extend $\mathcal{L}_c$ to $\mathcal{L}_c\cup\{0\}$. Then the same logical constraint should be expressed as an equality. This is because, then, one of these extended levels has to be selected. That is, the same logical constraint that at most one non-zero level of a control must be picked would be enforced by the following algebraic constraint:
$$\sum_{l\in\mathcal{L}_c\cup\{0\}}x_{cl}=1,\quad \forall c\in\mathcal{C}.$$
The introduction of this explicit auxiliary level of zero (with zero costs and zero efficacy) may help to simplify the optimization, as was the case for the multiplicative model in the paper. 
