I wish to show the following: Let $I$ be any finite interval (i.e. Bounded) of any type (open, closed, half open).

1) If $I_1,\dotso, I_n$ are also intervals, pairwise disjoint such that $$I=\bigcup_{k=1}^n I_k$$ Then $$b-a = \sum_{k=1}^n b_k -a_k$$ (Where these are the endpoints of the respective intervals)

I know this can be easily accomplished using additivity of Riemann integration and characteristic/indicator functions.

I wish to prove this from first principles. I noticed in a general semi ring setting, the measure is defined to have such a property and this is used to prove various inequalities involving subsets and unions (the general analogs of the two facts i list below) Instead here, we'd first prove

2) If $I_1,\dotso, I_n$ are also intervals, pairwise disjoint such that $$I\supset \bigcup_{k=1}^n I_k$$ Then $$b-a \geq \sum_{k=1}^n b_k -a_k$$


3) If $I_1,\dotso, I_n$ are also intervals, (not necessarily disjoint) such that $$I\subset \bigcup_{k=1}^n I_k$$ Then $$b-a \leq \sum_{k=1}^n b_k -a_k$$

And then 1) would easily follow from 2) and 3) together.

Note I only wish to do this for finite unions/sums. I know countably infinite versions of 3) will require Heine-Borel to reduce it to the finite case.

But I cannot finish the induction step for either 2) or 3). If anyone has any suggestions or hints, it'd be greatly appreciated. Thank you.


1 Answer 1


Edit: (Sorry, I misread your question)

The thing to notice here is that only case we care about is when we have two intervals:

Let $I_1=[a_1,b_1],I_2=[a_2,b_2]$ be disjoint with $I_1\cup I_2\subset I=[a,b]$ and assume (up to reordering) that $b_1\leq a_2$, then we know $a\leq a_1$, $b_1\leq a_2$, and $b_2\leq b$, so $b-a\geq b_2-a_1\geq b_2-(a_2-b_1)-a_1=(b_2-a_2)+(b_1-a_1)$.

Trivially, if $I_1\subset I$, then the length of $I_1$ is less than that of $I$. Now suppose we've shown that for $n$ disjoint intervals subset of some interval $I$, the sum of their lengths is less than the length of $I$. Now, given disjoint intervals $I_k=[a_k,b_k]$ with $1\leq k\leq n+1$ with $\bigcup I_k\subset I=[a,b]$, we can reorder and rename them to $I_1',\ldots I_{n+1}'$ such that $b_k\leq a_{k+1}$. Let $J_1=[a_1,b_n]$, and $J_2=[a_{n+1},b_{n+1}]$, then by our inductive hypothesis, $\sum^n (b_k-a_k)\leq b_n-a_1$, and from what we've shown above it follows that $$b-a\geq (b_n-a_1)+(b_{n+1}-a_{n+1})\geq\sum^n(b_n-a_n)+(b_{n+1}-a_{n+1}).$$

This shows (2). I think (3) should be in the same spirit.

  • $\begingroup$ Thank you for the answer but there may have been a misunderstanding. I wish to prove 1) from 2) and 3). I have only shown 1) via Riemann integration. I do not wish to use this result to prove 2) and 3). $\endgroup$ Dec 1, 2015 at 5:41
  • $\begingroup$ Oops you're right... $\endgroup$
    – Couchy
    Dec 1, 2015 at 20:19
  • $\begingroup$ Ahh yes thank you again. The reordering assumption was the key part I failed to see. I will work out (3) now. Much appreciated! $\endgroup$ Dec 1, 2015 at 20:30

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