Nonexistence of extension of length to all subsets of real numbers

[Genady]

I've started studying measure theory by this book: Measure, Integration & Real Analysis (Graduate Texts in Mathematics) (It is free on Kindle, btw.)

It pretty much begins with proving that it is impossible to simply extend the concept of length from a real interval to an arbitrary set of real numbers. The proof is a bit formal, and I want to make it more intuitive while still rigorous. Would like to hear if the following description is good and if it can be improved.

So, we want to have a real-valued, non-negative function 'length' defined for any set of real numbers with the following desired properties:

a) it gives length of b-a for an interval [a, b],

b) if set A is a union of disjoint sets A₁, A₂, etc., then length(A) = length(A₁) + length(A₂) + ...,

c) length of a set does not change if the set shifts as a whole along the real line right or left.

Let's assume that such function exists. Consider interval [0, 1]. It can be partitioned so that each number x belongs to a subset consisting of all numbers whose distance from x is a rational number. IOW, if y-x is rational then y and x are in the same subset. 

Let set V to contain exactly one element from each subset of the partition above. If we shift V by a rational distance r, we get a set V_r which contains different elements from the subsets. Since all elements in each subset are separated from each other by rational distances between 0 and 1, the union of all sets V_r which are shifted by all rational distances r between -1 and 1, covers the entire interval [0, 1]. So, the sum of lengths of all sets V_r is greater than or equal to the length of the interval [0, 1], i.e. greater than or equal to 1. Since every V_r is just a shifted V, they all have the same length, which is therefore greater than 0.

OTOH, all V_r are shifted from the original V by maximum 1 to the right or to the left. Thus, their union is covered by the interval [-1, 2] and therefore the length of the union is less than or equal to 3. By taking enough of the shifted sets V_r we can get the sum of their lengths to be greater than 3. Then we get disjoint sets with the sum of their lengths being greater than the length of their union. This contradicts the property b) above. Thus, a function with the properties a), b), and c) does not exist.

Is it clear enough? How can this explanation be improved?

Edited 22 hours ago by Genady

[Genady]

Notice that the axiom of choice has been used in the proof above.