Formally speaking, the Lebesgue-measure $\lambda$ on $\mathbb{R}$ is the restriction of the outer Lebesgue-measure $$ \lambda^*(A)=\inf\left\{\sum_{n=1}^{\infty}p(C_n): A\subset\bigcup_{n=1}^{\infty}C_n, \forall n\in\mathbb{N}: C_n\in\mathcal{S}\right\} $$ where $p$ is the Lebesgue pre-measure on the semiring $\mathcal{S}$ of all intervalls of the form $(a,b]$ defined by $p((a,b])=b-a$ and $p(\emptyset)=0$ on the $\sigma$-algebra $\mathcal{M}(\lambda^*)$ of all sets that are $\lambda^*$-measurable. One says that $\lambda$ is the unique continuation of $p$ and especially $\mathcal{B}(\mathbb{R})\subset\mathcal{M}(\lambda^*)$, i.e. all Borel sets are Lebesgue-measurable. So far so good.

Now I have one (maybe very stupid) problem.

If one has f.e. the intervall $(a,b)$ (which is in the Borel-$\sigma$-algebra and therefore formally it is Lebesgue-measurable) it is $$\lambda((a,b))=\lambda^*_{|\mathcal{B}(\mathbb{R})}((a,b))=b-a. $$

  1. Why?

  2. Is it right that calculating the Lebesgue-measure of a Borel-set is calculating the outer Lebesgue-measure in fact? So that writing $\lambda$ is in fact not very precisely? Instead one should write $\lambda^*_{|\mathcal{B}(\mathbb{R})}$.

  3. In which way is $\lambda=\lambda^*_{|\mathcal{B}(\mathbb{R})}$ a continuation of $p$, because $\lambda=p$ on $\mathcal{S}$?


1 Answer 1

  1. Because that is the definition of restriction: if we have a function $f\colon A\to B$ and a $C\subseteq A$, then $f|_C$ is the function $C\to B$ such that for all $c\in C$ we have $f|_C(c)=f(c)$. Unless you're asking why $\lambda^*((a,b))=b-a$ -- that follows from continuity of $p$ on $\emptyset$, for example.
  2. Why would that be imprecise? If you define $f(x)=x^5+3x^2+6x$, would you regard writing $f(1)=10$ (instead of $1^5+3\cdot 1^2+6\cdot 1=10$) as imprecise?
  3. I'm not sure what is meant by continuation, but I think it's most likely meant in a sense synonymous to "extension". In that case, yes, the point is, $p=\lambda|_{\mathcal S}$.
  • $\begingroup$ 1. Yes, my question is, because $\lambda^*((a,b))=b-a$, but now I think it is clear: $(a,b)=\bigcap_{n\geqslant 1}(a,b+\frac{1}{n}]$ and $\lambda^*(a,b+1]=b+1-a<\infty$ so it is $\lambda^*((a,b)=\lim_{n\to\infty}b+\frac{1}{n}-a=b-a$. 2. I do not see what your example has in common with that to be honest. I think it would be more clear not to write $\lambda$ but to write it as the restriction of an outer measure to make clear how the measure is constructed. 3. Yes, I mean extension. $\endgroup$
    – mathfemi
    Sep 16, 2014 at 12:24
  • 1
    $\begingroup$ @mathfemi: It is not really important how the measure is constructed: its important properties are that: it is a (regular, countably additive, positive) measure, it attains on intervals the values we want it to attain, and it is translation invariant. The rest is only really a construction to show that it exists. $\endgroup$
    – tomasz
    Sep 16, 2014 at 12:31

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