Why does not the Euclidean space support a countably additive measure defined for all subsets?

Question

The three axioms that a measure function $m$ need satisfy are as follows. Let the word "measure" mean the value returned by this measure function, $m$ defined on a set in $\mathbb{R^n}$.

1. The measure of an interval $I$ is its length: $m(I)= l(I)$;
2. The measure is translation invariant: $m(E+y) = m(E)$;
3. The measure is countably additive over countable disjoint union of sets: $$m(\bigcup_{k=1}^\infty E_k)=\sum_{k=1}^{\infty} m(E_k)$$.

My questions are: Why cannot a set function be defined that satisfy all the above? What is the geometry or structure of $\mathbb{R^n}$ that prevents this from happening?

Considering $\mathbb{R^n}$ is complete, totally ordered, it seems like a space that is not pathological at all, yet it harbors this problem when comes to measure. Could someone explain why? Thank you.

Also, even though existence of a non-measurable set can be shown, can one be actually constructed?