# Constructing the Lebesgue Measure

I am studying for an exam and I am trying to work out a concise construction of the Lebesgue measure. Here is what I have:

If $\mathcal{A}$ is an algebra of open intervals in $\mathbb{R}$ with a premeasure $\mu_0:\mathcal{A}\to[0,\infty)$ where $\mu_0(A)$ is the length of $A\in\mathcal{A}$. Now we want to define the outer measure $\mu^*:\mathcal{P}(\mathbb{R})\to[0,\infty]$ such that if $E\subset\mathbb{R}$, take $E\subset\bigcup_{i=1}^\infty A_i$ for $A_i\in\mathcal{A}$ for all $i$ such that $\mu^*(E)=\inf\sum_i\mu_0(A_i)$. We can show then that $\mu^*(\emptyset)=0$ and $\sigma$-additivity on $\mu^*$. Now, by Caratheodory, we have that $\mu^*$ is a complete measure in the $\sigma$-algebra of $\mu^*$ measurable sets, $\mathcal{M}$ (i.e. a set $K$ is $\mu^*$ measurable if $\mu^*(J)=\mu^*(J\cap K)+\mu^*(J\cap K^C)$ for all $J\subset\mathbb{R}$). But then $\mathcal{A}\subset\mathcal{M}$ which means $\mu^*$ is the unique completion of $\mu_0$, so $\mathcal{M}=\sigma(\mathcal{A})=\mathcal{B}_\mathbb{R}$. So, to give us what we want, we have a complete measure on $\mathcal{B}_\mathbb{R}$ that defines the length of intervals. This means the measure is the Lebesgue measure. And there we have it.

1. I know I didn't define $\mathcal{A}$ very specifically, is that ok?
3. I didn't use $F(x)=x$, which really defines the Lebesgue measure. I think I just found a way around that, but is it not ok to leave that out?
• For #1, by the topology axioms, the set of all sets which are open or closed is an algebra. At least in $\mathbb{R}$ it is the smallest algebra containing all open intervals. Thus if you want to extend your premeasure from this algebra, it should first be defined on this algebra. But the length of an interval is not well-defined on this entire algebra. So there is some work to be done to properly specify $\mu_0$. – Ian Oct 13 '15 at 15:33
• For #2, you should establish some uniqueness, so that the Lebesgue measure is the only possible complete extension of the premeasure you started with. (This is not trivial; as I recall you need the space to be $\sigma$-finite to get this uniqueness.) For #3, it is built into the definition of the premeasure. – Ian Oct 13 '15 at 15:36
• @Ian: The set of sets which are open or closed does not form an algebra, e.g. $(0,1) \cap [1/2, 2) = [1/2, 1)$ is neither open nor closed. The algebra in question here is (usually) just the set of all finite disjoint unions of intervals of the form $(a,b]$. – PhoemueX Oct 13 '15 at 15:48
• @PhoemueX Good catch. To me that raises an interesting question about construction of Borel measures (though I shouldn't distract from this post). Anyway, my main point in #1 was that you need to specify what $\mathcal{A}$ is and what $\mu_0$ is for all elements of $\mathcal{A}$, which allthatmath did not do in the original question. – Ian Oct 13 '15 at 15:54