Why it is not possible to construct a set function that is defined for all sets of real numbers with the following 3 properties? I was reading page 30 of Royden and Fitzpatrick" real analysis ", fourth edition  and the book said: " it is not possible to construct a set function that is defined for all sets of real numbers with the following 3 properties:
1- The measure of an interval is its length.
2- The measure is translation invariant.
3- The measure is countable additive over countable dis-joint unions of sets."  and then the book said as a justification for this that we should look at page 48. here is a part of pg. 48:



After that, on the remaining part of the page, the book started to prove thm. 17.
My question is:
What is on pg. 48 say that it is not possible to construct a set function that is defined for all sets of real numbers with the previous 3 properties? could anyone explain this to me, please?
 A: Suppose we have such a function $f$. Now let's use some choice set $\mathcal{C} = \mathcal{C}_E$, where we took $E=[0,1]$.
First, $\{\lambda + \mathcal{C}\}_{\lambda\in \mathbb{Q},\, 0\leq \lambda \leq 1}$ is countable and disjoint, so by property 3 we have
$$
f\left(\bigcup_{\lambda\in\mathbb{Q}, \, 0\leq \lambda \leq 1} (\lambda + \mathcal{C}) \right) = \sum_{\lambda \in \mathbb{Q},\,0\leq \lambda \leq 1} f(\lambda + \mathcal{C}).
$$
Since each point $x\in \mathopen[0,1\mathclose]$ is rationally equivalent to a point $c\in \mathcal{C}$, we have $[0,1] \subset \bigcup_{\lambda\in\mathbb{Q},\, 0\leq \lambda \leq 1} (\lambda + \mathcal{C}) \subset [0,2]$. But $f$ is nondecreasing, so our left hand side lies between $1$ and $2$ by property 1.
Now for the right hand side : by property 2, each $f(\lambda + \mathcal{C})$ is equal to $f(\mathcal{C})$. We have a countable sum of the same real number. Either $f(\mathcal{C})=0$, and then the RHS is zero ; or $f(\mathcal{C})>0$, which implies that the RHS is $+\infty$. Both cases lead to a contradiction.
