Proving the uniqueness property of Lebesgue measure 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!
 A: Using only what has been stated in Terence Tao's book, one can follow the following steps:


*

*Show that $m$ agrees with the elementary measure on elementary sets.

*This implies $m(E)\le m^*(E)$, where $m^*(E)$ is the outer Lebesgue measure.

*Then we prove that $m(\cup_{n \ge 1} B_n)=\sum_{n \ge 1}|B_n|$, where $B_n$ are almost disjoint boxes.

*Then $m$ agrees with Lebesgue measure on open sets, since open sets is a union of almost disjoint boxes.

*Given this, we see that $m$ also agrees with Lebesgue measure on compact sets.

*Then we use the property of finite measurable set can be approximated below by a compact set to show that $m$ argees with Lebesgue measure on bounded measurable sets.

*This implies $m$ also agrees with unbounded measurable sets.

A: Start with a measure $m$ on the field of finite disjoint unions of intervals satisfying the stated properties. Note that this measure is $\sigma$-finite, which follows from properties (3) and (4) [just look at the intervals of unit length covering the whole space]. You can extend $m$ to an outer measure $m^*$ and then restrict it to the $\sigma$ field of Lebesgue measurable sets, calling it $m'$. $\sigma$-finiteness of $m'$ ensures uniqueness of measure on the $\sigma$ field of Lebesgue measurable sets. The Lebesgue measure already is a measure satisfying all these properties in the $\sigma$ field of Lebesgue measurable sets. Therefore, $m'=\lambda$. 
A: First assume that $E$ is bounded.
As mentioned above, one shows that $m$ agrees with elementary measure on elementary sets and hence $m(E)\le \lambda^*(E)$, where $\lambda$ is the Lebesgue measure.
Now let $A \supset E$ be an elementary set. By the Carathéodory criterion, $m(A) = \lambda(A) = \lambda^*(E) + \lambda^*(A \setminus E)$. In particular, $\lambda^*(E) = m(A) - \lambda^*(A \setminus E) \leq m(A) - m(A \setminus E) = m(E)$, by finite additivity. This shows that $m$ agrees with Lebesgue measure on bounded measurable sets.
Finally, if $E$ is unbounded, we can express $E$ as countable union of disjoint bounded segments, then the conclusion follows from countable additivity and the bounded case.
