Let $\Omega = (0,1]$, $\mathcal{S}=\{(a, b] : 0 \leq a \leq b \leq 1\}$
Define on $\mathcal{S}$ the function by $\lambda : \mathcal{S} \mapsto[0,1]$ by $$\lambda(\emptyset)=0, \quad \lambda(a, b]=b-a$$
How to show $\lambda$ is $\sigma$-additive? I can only show that it is finite additive.Any hint?
We exploit the Zeno's paradox and the Heine Borel theorem.
Suppose that $\{(a_i,b_i]\}_{i=1}^\infty\subset\mathcal S$ is disjoint and its union equals $(a,b]\in\mathcal S$. We clearly have $\sum_{i\leq N}(b_i-a_i) \leq b-a$ for all $N$, so $\sum_{i}(b_i-a_i) \leq b-a$. To prove the reverse inequality, let $\varepsilon>0$. Define $$a_i'=a_i-2^{-(i+1)}\varepsilon,\ b_i'=b_i+2^{-(i+1)}\varepsilon.$$ The intervals $(a_i',b_i')$ cover the compact interval $[a+\varepsilon,b]$, so finitely many of them, say $(a_1',b_1'),\cdots,(a_n',b_n')$, covers $[a+\varepsilon,b]$, and therefore $b-(a+\varepsilon)\leq\sum_{i\leq n}(b_i'-a_i')$. But now $$\sum_{i\leq n}(b'_i-a_i')\leq\sum _i(b_i-a_i+2^{-i}\varepsilon)\leq\sum_i (b_i-a_i)+\varepsilon,$$so $b-a\le\sum_i(b_i-a_i)+2\varepsilon$. Since $\varepsilon$ was arbitrary, the claim follows.