[Math] Proving the uniqueness property of Lebesgue measure

lebesgue-measuremeasure-theoryreal-analysis

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:

  1. Show that $m$ agrees with the elementary measure on elementary sets.
  2. This implies $m(E)\le m^*(E)$, where $m^*(E)$ is the outer Lebesgue measure.
  3. Then we prove that $m(\cup_{n \ge 1} B_n)=\sum_{n \ge 1}|B_n|$, where $B_n$ are almost disjoint boxes.
  4. Then $m$ agrees with Lebesgue measure on open sets, since open sets is a union of almost disjoint boxes.
  5. Given this, we see that $m$ also agrees with Lebesgue measure on compact sets.
  6. 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.
  7. This implies $m$ also agrees with unbounded measurable sets.
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