Probability distribution/measurment problem

The onedimensional measure problem can described by the following requirements: Is there an illustration $$ \mu_n: {\mathcal P} ({\mathbb R}^n) \ rightarrow [0, \infty] $$ with the following properties:
 * Positivity: $$ \mu_n (A) \ge 0 $$ for all $$ A \subset {\mathbb R} $$
 * Translation Invariance: $$ \mu(A) \, = \, \mu(A_v) $$ for all $$v \in {\mathbb R}$$ and $$A_v:=v+A:=\{v+a\,|\,a \in A\}$$.
 *  Normality : $$ \ mu([0,1]) \, = \, 1 $$,
 * $$ \sigma $$-Additivity: $$ \mu (\bigcup_{i = 1}^\infty A_i) = \sum_{i = 1}^\infty \mu (A_i ) $$ if $$ A_i \cap A_j = \emptyset $$ for $$ i \not= j $$?

Unsolvable of Measurement Problem
The onedimensional measurement problem cannot be solved on the power set on $$\mathbb{R}$$ as shown by Giuseppe Vitali in 1905, who was the first to give an example of a non-measurable subset of real numbers, see Vitali set. His covering theorem is a fundamental result in measure theory.

The onedimensional measurement problem can easily be extended to $$n$$-dimensional spaces; this encompasses the dimensions 1, 2 and 3, that are accessible to our spatial perception, whereby
 * dimension 1 refers to length,
 * dimension 2 to an area in a plane and
 * dimension 3 to measurment of the volume.

Due to the fact, that the measurement problem for $$\mathbb{R}$$ cannot be solved as shown by Vitali, the sigma-algebra was introduced as the domain of the measure $$\mu$$, that replaces especially the power set as domain of the probability measure $$P$$ on a $$\sigma$$-algebra $$\mathcal{S}$$.

n-dimensional Case of the Measurement Problem
The measurement problem in the $$n$$-dimensional case is:

Is there an measure $$ \mu_n: {\mathcal P} ({\mathbb R}^ n) \rightarrow [0, \infty] $$ with the following properties:
 * Positivity: $$ \mu_n (A) \ ge 0 $$ for all $$ A \subset {\mathbb R}^n $$ (this condition is already in the default of the image set of the figure),
 * Congruence: $$ \mu_n (A) \, = \, \mu_n (B) $$ if A and B are congruent,
 * Normality: $$ \ mu_n ([0,1] ^ n) \, = \, 1 $$,
 * $$\sigma$$-Additivity: $$ \mu_n (\bigcup_{i = 1}^\infty A_i) = \sum_{i = 1}^\infty \mu_n (A_i ) $$ if $$ A_i \cap A_j = \emptyset $$ for $$ i \not= j $$?