Is it possible to construct an uncountable partition $(A_\delta)_{\delta\in[0,1]}$ of the unit interval $[0,1]$ with the following properties:

1. $\cup_{\delta\in[0,1]} A_\delta=[0,1]$;

2. If $\delta_1\neq \delta_2$, then $A_{\delta_1}\cap A_{\delta_2}=\emptyset$;

3. For each $\delta$ we have $\mu (A_\delta)=1$, where $\mu$ stands for the outer Lebesgue measure.

My motivation is as follows. Using Vitali sets of outer measure $1$, it is possible to construct a countable partition of $[0,1]$ with such properties. Indeed, one just need to put  $$A_\delta:=V+\delta \mod 1,\quad \delta\in\mathbb{Q},$$ where $V$ is the Vitali set of outer measure $1$.

My question is whether it is possible to construct an uncountable partition of $[0,1]$ with these properties.

Thanks!

Is it possible to construct an uncountable partition $(A_\delta)_{\delta\in[0,1]}$ of the unit interval $[0,1]$ such that $\mu (A_\delta)=1$ for each $\delta\in[0,1]$? ($\mu$ stands for the outer Lebesgue measure.)

My motivation is as follows: Using Vitali sets of outer measure $1$, it is possible to construct a countable partition of $[0,1]$ with such properties. Indeed, one just need to put 

$$A_\delta:=V+\delta \mod 1,\quad \delta\in\mathbb{Q},$$ where $V$ is the Vitali set of outer measure $1$.

My question is whether it is possible to construct an uncountable partition of $[0,1]$ with these properties.