in our lecture it was stated that the Lebesgue measure can be uniquely extended from a semiring to a sigma algebra by Caratheodory's theorem. Unfortunately, we did not show that it is unique on the semiring of the intervals $(a,b]$. My question is: Is it obvious that the Lebesgue measure defined by (i) $\lambda((0,1])=1$
(ii)$\lambda$ invariant under translations
(iii) $\lambda$ is $\sigma-$additive
is unique on the semiring or is this a very long proof?
If anything is unclear, please let me know.
The semiring of intervals $(a,b]$ is not closed under infinite disjoint unions, so I would change (iii) by $\lambda$ being finitely aditive and non-negative. I'll use $\sum$ and $+$ for disjoint union. Let $\lambda$ satisfy (i), (ii) and (iii). Notice that $$1=\lambda(0,1]=\lambda\left(\sum_{i=1}^n\left(\frac{i-1}{n},\frac{i}{n}\right]\right)=\sum_{i=1}^n\lambda\left(\frac{i-1}{n},\frac{i}{n}\right]=\sum_{i=1}^n\lambda\left(0,\frac{1}{n}\right]=n\lambda\left(0,\frac{1}{n}\right],$$ so $\lambda\left(0,\frac{1}{n}\right]=\frac{1}{n}$. $n\in\mathbb{N}$. Using translation invariance and finite aditivity, you can show that $\lambda(0,q]=q$ for every $q>0$ rational.
Now, let $a>0$. For every rationals $q,r>0$ satisfying $q<a<r$, we have
$$q=\lambda(0,q]\leq\lambda(0,a]\leq\lambda(0,r]=r$$
Taking the supreme for $q$ and the infimum for $r$, we obtain $\lambda(0,a]=a$. Using translation invariance, we obtain $\lambda(a,b]=b-a$ for every $a,b\in\mathbb{R}$.