To prove this, I claim that if $G= \cup_{k=1}^n(a_k, b_k)$,i.e., finite union of disjoint open intervals. Then, $|A\cup G|\ge |A|+|G|$.

Proof: WLOG, suppose that 1) $a_k<b_k$ and that $a_1<a_2<...<a_n$. 2) $a_i,b_i\not\in A$.

For 2), indeed if $X\subset \{a_1,b_1,\cdots, a_n,b_n\}, X\subset A$, then $|A\cup G|=|(A-X)\cup G|, $ because $X$ is a finite set. Therefore, we could consider $A-X$ instead of $A$. So the assumption 2) is valid.

Take any sequence of open intervals $\{I_n\}$ that cover $A\cup G$,i.e., $A\cup G\subset \cup_n I_n$.

Define $K_n, L_n, J_n, M_n$ as follows:

$J_n=(-\infty, a_1)\bigcap I_n; K_n= G\bigcap I_n, L_n= (\bigcup_{k=1}^{n-1}(b_k,a_{k+1}))\cap I_n, M_n=(b_n, \infty)\bigcap I_n$.

Clearly, $J_1,L_1,M_1, J_2, L_2, M_2,\cdots$ cover $A$ and $\{K_n\}$ covers $G$.

Since, $l(I_n)= (l(J_n)+l(L_n)+l(M_n))+l(K_n)$, we have $$\sum l(I_n)= \sum(l(J_n)+l(L_n)+l(M_n))+\sum l(K_n)\ge|A|+|G|$$

Taking infimum over all such open intervals $\{I_j\}$, we get $|A\bigcup G|\ge |A|+|G|.$

This proves the claim. By subadditivity, we even have $|A\bigcup G|= |A|+|G|$.

Now coming back to the statement in title, it is known that every open set in $R$ is a countable disjoint union of open intervals so there exist $\{I_n\}$ such that $G=\bigcup_n I_n$.

By subadditivity, we have $|A\cup G|\ge |A\bigcup (\cup_{n=1}^k I_n)$ for every $k\in \mathbb N$.

By the claim proven above, $|A\cup G|\ge |A|+\sum_{n=1}^kl(I_n)$ for every $k\in \mathbb N$.

Letting $k\to \infty$, $|A\cup G|\ge |A|+\sum_{n=1}^\infty l(I_n)\ge|A|+|G|.$

By subadditivity, $|A\cup G|\le |A|+|G|$ so we get the equality. $$\tag*{$\square$}$$

My question: In theorem 2.62 of Axler's measure theory book, Axler first proves the above theorem in case $G= (a,b)$ and then concludes using induction on $m$, the no. of open intervals (disjoint) whose union equals to $G$. Then, the conclusion is made for any open set $G$. But here, I have not used induction anywhere above so I want to know if induction is required here or not.


  • $\begingroup$ Are you suggesting that to demonstrate the claim for generic $G$ it suffices to demonstrate it for finite intervals? $\endgroup$
    – FShrike
    May 18 at 16:51
  • $\begingroup$ (using machinery, the result follows for open $G$ because open sets are Lebesgue measurable : ) ) $\endgroup$
    – FShrike
    May 18 at 16:54
  • $\begingroup$ @FShrike: The book first proves the result for G being a union of finitely many m disjoint open intervals by first proving the result for G= (a,b), single interval,i.e., m=1. Then, the book suggests by using induction on m. My question is: in my attempted proof above, I nowhere had to use induction. So is it correct or did I miss something? $\endgroup$
    – Koro
    May 18 at 17:15
  • $\begingroup$ @FShrike: I understand why proving for finite is enough to prove for any open G disjoint with A. $\endgroup$
    – Koro
    May 18 at 17:24
  • $\begingroup$ I now see why it follows for general open sets given the conclusion for finitely many disjoint intervals, took me a moment. Induction seems very easy, not sure why you want to avoid that $\endgroup$
    – FShrike
    May 18 at 17:36

1 Answer 1


In your notation for $G$, instead of using $n$ for the number of open intervals that make up $G$ it might be better to use $m$ (as Axler does) to avoid confusion with the notation for the open interval cover $\{ I_n \}$ of $A \cup G$. When you say $\{K_n\}$ covers $G$ it is not strictly correct to immediately conclude that $\sum l(K_n) \geq |G|$ because the $K_n$'s are not open intervals, but are finite unions of open intervals, so that $l(K_n)$ is not well defined. Axler [1, p14] only defines the interval length function $l$ for open intervals (though it would be a simple matter to extend that definition to all interval types) - and his definition of Outer Measure (§2.2 on p14) is given in terms of this function $l$.

To make your proof a formal proof you would have to write something like :

$$ K_n = P^{(n)}_1 \cup \cdots \cup P^{(n)}_{r_n} $$

where $P^{(n)}_1, \ldots, P^{(n)}_{r_n}$ are disjoint open intervals and each $r_n \geq 0$.

Then as in your method, considering how the length of $I_n$ is split up by the various open subintervals you would have : $$ l(I_n) = (l(J_n) + l(L_n) + l(M_n)) + \sum_{j = 1}^{r_n} l(P^{(n)}_j) $$

then the definition of Outer Measure applied to $G$ would allow you to say :

\begin{eqnarray} \sum_{n = 1}^{\infty} \; \sum_{j = 1}^{r_n} \; l(P^{(n)}_j) \geq |G| \label{eq:double-series} \tag{1} \end{eqnarray}

since the countable collection of open intervals $\{P^{(n)}_j\}$ covers $G$, leading to :

$$ \sum_{n = 1}^{\infty} l(I_n) \geq |A| + |G|. $$

So the induction method is not needed and with the above modification your method works. However the induction method on $m$ given in Axler is very simple - we just 'break off' one open interval at a time, and we obtain a much clearer and less cumbersome proof than the above. The induction step is very immediate so there is no advantage in obviating the induction.

It is worth noting in (1) that we are actually invoking a theorem regarding double infinite series, since we are rearranging a single infinite series (namely a series as in Axler's definition in [1, §2.2, p14]) by partitioning it into a collection of sub-series (in this case the sub-series are all finite sums). Some useful theorems on double series and series rearrangements are described in this answer to question Rearrangements of absolutely convergent series (we use Theorem 1 in that answer to infer (\ref{eq:double-series}) above).


[1] Sheldon Axler (2020), Measure, Integration & Real Analysis, Springer Graduate Texts in Mathematics, https://measure.axler.net/.


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