Lebesgue measure on intervals

Question

We know that Lebesgue measure is the standard way of assigning a length, area or volume to subsets of Euclidean space (e.g., the Lebesgue measure of the interval $[a,b]$ of real numbers is the length $b-a$). We consider $l_1=(0,1)$ and the Lebesgue measure $\mathcal{M}$ on it. Can we replace $(0,1)$ with $l_2=[0,1]$? (i.e., Is the Eq. $\mathcal{M}(l_1)=\mathcal{M}(l_2)$ correct?). Why?

Answer 1

Let us notice that the Lebesgues measure on $(0,1)$ (resp. $[0,1]$) is translation-invariant and measures closed subintervals which means that for $[a,b]\subset (0,1)$ (resp. $[a,b]\subset [0,1]$) and $t>0$ such that $[a,b]+t=[a+t,b+t]\subset (0,1)$ (resp. $[a+t,b+t]\subset [0,1]$) then $\mu([a,b])=\mu([a,b]+t)$. Let us call this class of measures (with $\mu((0,1))<+\infty$, resp. $\mu([0,1])<+\infty$) T-invariant, with this at hand one can show that, for these measures,

Every singleton has measure zero.

Indeed, take two singletons $\{a\},\{b\}$ with $a<b$, using $t=b-a$, you can see that $\mu(\{a\})=\mu(\{b\})$. Now for every $a\in (0,1)$, the set $S_a=\{\frac{a}{n}\}_{n\geq 1}$ is measurable and $$ \mu(S_a)=\sum_{n\geq 1}\mu(\{\frac{a}{n}\})\leq \mu((0,1))<+\infty \qquad (resp.\ \leq \mu([0,1])<+\infty) $$ but, due to the fact that $\mu(\{\frac{a}{n}\})=\mu(\{a\})$ for all $n$, one has $\mu(\{a\})=0$ (for $0$ or $1$, you see that $\mu(\{0\})=\mu(\{1\})=\mu(\{a\})=0$ for any $a\in (0,1)$)

So the only way to extend a T-invariant measure $\mu$ in $(0,1)$ as a T-invariant one of $[0,1]$ is to add $\mu(\{0\})=\mu(\{1\})=0$.

Hope it helps.