Why is there no probability measure $\mu:2^{[0,1]}\to [0,1]$ on $[0,1]$ s.t. $\mu([0,x])=x$? Why is there no probability measure $\mu:2^{[0,1]}\to [0,1]$ on $[0,1]$ s.t. $\mu([0,x])=x$  and $\mu$ is invariant under translation ?

Attempts
Let $\mathcal B([0,1])$ the Borel set of $[0,1]$. I know that on $\mathcal B([0,1])$, there is a unique measure s.t. $m([0,x])=x$, it's the Lebesgue measure. We can extend $m$ on $\mathcal M([0,1])$ the set of Lebesgue measurable set on $[0,1]$. 
Question: Now, why there are no measures $\mu:2^{[0,1]}\to [0,1]$ s.t. $\mu(A)=m(A)$ for all $A\in \mathcal M([0,1])$ ? 
I know that there are always $A,B$ s.t. $m^*(A\cup B)<m^*(A)+m^*(B)$. So, I think I should prove that if such $\mu$ exist, then there is an exterior measure $\mu^*:2^{[0,1]}\to [0,1]$ and $\mu^*=m^*$, but I failed to prove it. 
My theorem is the following one

 A: Who says there is no such measure?  
Certainly if $\mathfrak c$ is not a real-valued measurable cardinal, then there is no such measure.  (For example, if CH holds.) 
But it is believed that "$\mathfrak c$ is a real-valued measurable cardinal" is consistant with ZFC; in that case there is such a measure.  
See https://en.wikipedia.org/wiki/Measurable_cardinal#Real-valued_measurable

So ... Assume $\mathfrak c$ is a real-valued measurable cardinal.  That is: there is a probability measure $\mu : \mathcal P(\mathbb R) \to [0,1]$, where $\mathcal P(\mathbb R)$ is the power set of $\mathbb R$.  
It is known (ref?) that $\mathfrak c$ is not a $2$-valued measurable cardinal.  So the nondecreasing map $\varphi : \mathbb R \to [0,1]$ defined by $\varphi(t) = \mu\big((-\infty,t]\big)$ has no jumps, thus $\varphi$ is continuous.  By the intermediate value theorem, $\varphi$ maps $[-\infty,+\infty]$ onto $[0,1]$.  
Define the "image measure" $\mathbb P = \varphi_*(\mu)$ as usual.  That is, for $E \subset [0,1]$ let $\mathbb P(E) = \mu\big(\varphi^{-1}(E)\big)$.  Then $\mathbb P$ is a probability measure defined on $\mathcal P([0,1])$ with the property $\mathbb P\big([0,t]\big) =t$ for $0 \le t \le 1$.
