# Lebesgue Premeasure via Transfinite Induction

If $I=[a,b)$ we write $|I|=b-a$ for the length of $I$. Given a theorem of Caratheodory, the tricky part in showing the existence of Lebesgue measure is this:

Lemma If $[0,1)$ is the disjoint union of a countable collection $(I_j)$ of half-open intervals then $\sum_j|I_j|=1$.

It's easy to conclude that this is easier than it really is, if, as for example in Easy proof for existence of Lebesgue-premeasure, one assumes that we can order the $I_j$ in such a way that $I_{j+1}$ lies to the right of $I_j$. The actual arrangement of the $I_j$ within $[0,1)$ can be much more complicated than that. Let's call the failed argument the "naive" proof.

The lemma is more or less Lemma 1.15 in Folland Real Analysis (Folland does it for general Lebesgue-Stieltjes measures). Folland calls the proof "remarkably fussy", which seems right. He says that the reason for the remarkable fussiness is that the arrangement of the intervals within $[0,1)$ can be order-isomorphic to any countable ordinal.

His point is that a countable ordinal can be complicated. I read this comment once and said to myself a countable ordinal is not that bad. In fact one can fix the naive proof, using a simple transfinite induction. (And we don't need to actually mention ordinals per se at all.)

Question Anyone ever seen the proof below?

Proof of the Lemma Given two disjoint intervals $I$ and $J$, write $I<J$ if $x<y$ for every $x\in I$ and $y\in J$.

Suppose that $[0,1)$ is the disjoint union of the half-open intervals $I_j$. First we note that the $I_j$ are well-ordered (under the order on intervals we just defined).

Indeed, if we had a decreasing sequence of $I_j$ they would accumulate at some point $x$. But there exists $k$ so $x\in I_k=[a_k,b_k)$, and hence disjointness shows our sequence of intervals cannot accumulate at $x$ after all.

Just for the sake of notation, we want there to be a rightmost interval. So add the interval $[1,2)$ to the collection, so now the union is $[0,2)$. The letters $I$ and $J$ will always refer to intervals in our collection.

We claim that for every $J$ we have $$\left|\bigcup_{I\le J}I\right|=\sum_{I\le J}|I|\quad(*).$$

If not, suppose that $J$ is the smallest counterexample: For every $J'<J$, ($*$) holds with $J'$ in place of $J$. Then $$\bigcup_{I<J}I=\bigcup_{J'<J}\left(\bigcup_{I\le J'}I\right),$$so $$\left|\bigcup_{I<J}I\right|=\sup_{J'<J}\left|\bigcup_{I\le J'}I\right| =\sup_{J'<J}\sum_{I\le J'}|I|=\sum_{I<J}|I|,$$and we're done (add $J$ to the union on the left and $|J|$ to the sum on the right).

So ($*$) holds for every $J$, and setting $J=[1,2)$ shows precisely that $\sum|I_j|=1$.

• Drive-by downvote - one wonders what the problem is... Commented Aug 25, 2015 at 0:46
• the problem is that he has a much simpler proof than the book and wants to know if he's right . looks ok to me. Commented Aug 25, 2015 at 1:17
• @user254665 I think you missed who wrote what. I'm the OP and I also wrote that comment. I wasn't wondering what the OP's problem was - got no problem, I know it's right. I got a downvote - I was wondering what problem the downvoter saw here. Commented Aug 25, 2015 at 1:21
• Nice post. Maybe it would be worthwhile to point out that completeness of $\Bbb{R}$ is used in an essential way to show that your intervals are well-ordered. Because my usual way of showing that "alternative" ways of proving countable additivity do not work is that they do not use completeness and thus would also work for $\Bbb{Q}$ instead of $\Bbb{R}$ (with a suitable interpretation of intervals etc.). Commented Aug 25, 2015 at 8:16
• I also came up with this during my second real analysis class. Transfinite recursion is indeed very intuitive. But could you explain the first supremum in the third from last line? My thought was to discuss the successor case and limit case separately, and in the limit case take a countable cofinal sequence and use $\sigma$-additivity. Commented May 1, 2022 at 2:52