# Show this function is $\sigma$-additive

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?

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.
• We only know the length of interval $(a,b]$ but otherwises,how can you prove that $b-a\leq\sum_{i\leq n}(b_i'-a_i')$ which involves other type of interval $(x,y)$ and $[x,y]$? – Spaceship222 Aug 14 at 4:47
• Sure enough, $\lambda$ is only defined for left half-open intervals, but I presume that you know what the length of an interval is, and it shouldn't be difficult to prove that if finitely many open intervals cover an interval, then the total length of the covering interval is at least the length of the interval being covered. – Ken Aug 14 at 5:45
• For $b-(a+\varepsilon)\leq\sum_{i\leq n}(b_i'-a_i')$, I try to prove it like this: let $b_M = \max\{b_i'\}$, $a_m = \min\{a_i'\}$, then $(a+\varepsilon,b] \subset (a_m, b_M]$,so $b-a-\varepsilon \le b_M - a_m \le \sum_{i\leq n}(b'_i-a_i')$ if we only choose $(a_i' , b_i') \cap [a+\varepsilon,b] \ne \emptyset$. Are I right? – Spaceship222 Aug 15 at 12:24
• @Spaceship222 Yes you are, although the argument that is necessary to prove the inequality $b_M-a_m\leq \sum _{i\leq n}(b'_i-a_i')$ is essentially the same as the one for proving $b-a-\varepsilon \leq \sum _{i\leq n}(b'_i-a_i')$. If you know Riemann integral, then you could also integrate the functions $\chi _{[a+\varepsilon, b]}$ and $\sum_{i \leq n} \chi_{(a_i',b_i')}$ to obtain the desired inequality. (Here $\chi_E$ denotes the characteristic function of $E\subset \mathbb R$.) – Ken Aug 15 at 23:54