Lemma 1.2.15 (The measure axioms)

The first claim is trivial.

The second claim we prove by generalizing cases.

• $E_n$’s are compact.
  We show that the left hand side is both greater than equal and less or equal to the right hand side.

  The first direction is trivial, and follows directly from the countable subadditivity property of the Lebesgue measure (Exercise 1.2.3):

  $$m\left(\bigcup _{n=1}^{\infty }{E}_n\right)\le \sum _{i=1}^{\infty }m(E_n)$$

  Now we prove the other direction. Notice, that by Exercise 1.2.4 compact disjoint sets $E_n$ must be separated by some set distance; by Lemma 1.2.5 we thus have the finite version of the claim: for any $E_1,\dots ,E_N$ it is true that $m\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{E}_n\right)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{n}m(E_n)$. By monotonicity of the Lebesgue measure

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}:m\left(\bigcup _{n=1}^{\infty }{E}_n\right)\ge m\left(\bigcup _{n=1}^N{E}_n\right)=\sum _{i=1}^{N}m(E_n)$$

  By monotonicity of limits, we thus obtain by taking $N\to \infty$ from both sides:

  $$m\left(\bigcup _{n=1}^{\infty }{E}_n\right)\ge \sum _{i=1}^{\infty }m(E_n)$$

• $E_n$’s are bounded.
  The trick is to make $E_n$’s look like they are disjoint; by Exercise 1.2.7 (v) we can replace each $E_n$ by a closed (and bounded) set $K_n\subseteq E_n$ such that $m(E_n∕K_n)=m(E_n)-m(K_n)\le 𝜖∕2^n$. Again, the first direction is trivial and follows directly from the countable additivity of Lebesgue measure

  $$m\left(\bigcup _{n=1}^{\infty }{E}_n\right)\le \sum _{i=1}^{\infty }m(E_n)$$

  On the other hand, employing the $𝜖∕2^n$ and the monotonicity of series we deduce

  $$\sum _{n=1}^{\infty }{E}_n\le \sum _{n=1}m(K_n)+𝜖=m\left(\bigcup _{n=1}^{\infty }{K}_n\right)+𝜖\le m\left(\bigcup _{n=1}^{\infty }{E}_n\right)+𝜖$$

  where we have used the additivity property applied to the compact sets. The arbitrary nature of the choice of $𝜖$ implies

  $$\sum _{n=1}^{\infty }{E}_n\le m\left(\bigcup _{n=1}^{\infty }{E}_n\right)$$

  as desired.

• General case
  Now we do not assume that $E_n$ are bounded any more. The basic idea is to decompose each $E_n$ as a countable disjoint union of bounded Lebesgue measurable sets. First, notice that we can decompose ${ℝ}^d$ as (disjoint) overlapped annuli of growing size:

  $${ℝ}^d=\bigcup _{m=1}^{\infty }{A}_n:=\bigcup _{m=1}^{\infty }\left\{x\in {ℝ}^d:m-1\le \parallel x\parallel <m\right\}$$

  Then each $E_n$ is the countable disjoint union of the bounded measurable sets $E_n\cap A_m$, i.e.,

  $$m(E_n)=m\left(\bigcup _{m=1}^{\infty }{E}_n\cap A_m\right)=\sum _{m=1}^{n}m(E_n\cap A_m)$$

  and thus, we apply the countable additivity for bounded sets:

  $$\begin{array}{llll}\hfill m\left(\bigcup _{n=1}^{\infty }{E}_n\right)& =m\left(\bigcup _{n=1}^{\infty }\bigcup _{m=1}^{\infty }{E}_n\cap A_m\right)=m\left(\bigcup _{(n,m)\in {\mathbb{N}}^2}{E}_n\cap A_m\right)& \hfill & \\ \hfill & =\sum _{(n,m)\in {\mathbb{N}}^2}m\left(E_n\cap A_m\right)=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }m(E_n\cap A_m)& \hfill & \\ \hfill & =\sum _{n=1}^{\infty }m(E_n)& \hfill & \end{array}$$

  where in the last steps we obviously used the Tonelli’s theorem.