I'm having trouble in showing the following statement: The Lebesgue measure is the only map $E \to m(E)$ from the class of measurable sets to $[0,+\infty]$, which satisfies the following properties:
$(i)$ Empty set property: $m(\phi)=0$
$(ii)$ Countable additivity: For disjoint measurable sets $E_1,E_2,\ldots$ ; $$m\bigg(\bigcup_{n=1}^{\infty}E_n\bigg)=\sum_{n=1}^{\infty}m(E_n)$$
$(iii)$ Translation invariance: For any measurable set $E \subset \mathbb{R}^d$ and for any $x \in \mathbb{R}^d$, $E+x := \{y+x|y \in E\}$ is measurable and $m(E+x)=m(E)$.
$(iv)$ Normalization: $m([0,1]^d)=1$, i.e. measure of the unit hypercube is $1$.
My approach: To go via Lebesgue outer measure. We already know that the only mapping from the class of all subsets of $\mathbb{R}^d$ (which obviously contains the class of all measurable sets in $\mathbb{R}^d$) to $[0,+\infty]$, satisfying $(i)$, a weaker version of $(ii)$ (only subadditivity) and $(iv)$, is the Lebesgue outer measure $m^*(\cdot)$. Then the job left is to show that with strict additivity and translation invariance, the Lebesgue outer measure $m^*(\cdot)$ is upgraded to the Lebesgue measure $m(\cdot)$, which I cannot complete. Any help is greatly appreciated!
Best Answer
Using only what has been stated in Terence Tao's book, one can follow the following steps: