# Prove that $\sum_{j=1}^{n} \mathbb{P}(I_j) \geq \mathbb{P}(I)$, when $\bigcup_{j=1}^n I_j \supseteq I$.

Let $I_1, I_2, \cdots, I_n$ be a finite collection of intervals on $[0,1]$, whose union contains an interval $I$, then $$\sum_{j=1}^{n} \mathbb{P}(I_j) \geq \mathbb{P}(I),$$ where $\mathbb{P}$ denotes the length of an interval.

I posted a proof of this theorem as a Lemma to a more general theorem, but my arguments seemed to confuse people. I have tried to rewrite it and tidy things up a bit, but it still seems convoluted, and I’ve struggled to excise the sloppy notation. Is there a simpler way to prove this theorem? Also, any advice on how to clean this proof up (and correct it if anything is still wrong) would be greatly appreciated!

Proof:

For any $1\leq j \leq n$, let $a_j$ be the left end point of $I_j$ and $b_j$ be the right end point of $I_j$. The sets $\{a_j\}$ and $\{b_j\}$ are finite and thus can be ordered. Let $\{a_{l_i}\}$ and $\{b_{k_i}\}$ be the ordered sets of these end points (ordered from least to greatest).

For the interval $I$ we have $$\bigcup_{j=1}^n I_j \supseteq I.$$

Letting $a$ and $b$ denote the left and right end points of $I$, respectively, without loss of generality we assume $$a_{l_1}\leq a\leq b_{l_1},$$ $$a_{k_n}\leq b\leq b_{k_n}.$$

The above simply says that $a$ and $b$ are in the intervals with smallest left end point and largest right end point, respectively (this could of course be the same interval). Using the ordering of end points, we can pair them up, from smallest to largest:

$$\sum_{j=1}^{n}(b_j - a_j)= \sum_{i=1}^{n}(b_{j_i} - a_{j_i})= \Big[\sum_{i =2}^{n-1}(b_{j_i} - a_{j_i}) +(b_{j_1} -a_{j_n})\Big]+(b_{j_n} -a_{j_1}) .$$

In the last expression, we have removed the smallest left end point and largest right end point from the sum. Since we assume some part of the interval $I$ is in every $I_n$, the remaining end points of intervals must overlap, which means, for any $i$, $$b_{j_{i-1}} \geq a_{j_i}.$$ Thus, we can reorder the sum again to show that

$$\sum_{i =2}^{n-1}(b_{j_i} - a_{j_i}) + (b_{j_1}-a_{j_n})= \sum_{i=2}^{n} (b_{j_{i-1}}- a_{j_i}) \geq0.$$

Returning to our original sum, it follows that

$$\sum_{j=1}^n (b_j - a_j) \geq b_{k_n} - a_{l_1} \geq b-a,$$

which implies $$\sum_{j=1}^{n} \mathbb{P}(I_j) \geq \mathbb{P}(I). \quad \square$$

For any finite collection of measurable sets $I,I_1, I_2,...I_n$ of the measure space $(\Omega, \mathscr A, P)$

$$\sum_{j=1}^n P(I_j)\geq P\left(\bigcup_{j=1}^n I_j\right)\geq P(I)$$

if $\bigcup_{j=1}^n I_j\supseteq I.$

EDIT

I think the following argumentation will help.

Finite unions of intervals behave like disjoint unions of intervals. This can be proven by mathematical induction. For two intervals, the statement is trivial. (The two intervals are either disjoint, containing, or overlaping...)

If the statement is true for $n$ intervals then...

• This assumes that $\mathbb{P}$ is a measure—which is what I’m trying to prove. Jul 18, 2018 at 12:45
• @MoedPolBollo: The class of intervals of $[0,1]$ and the interval length behave like a measure on the class of finite unions of the intervals. Every finite union of intervals can be considered as a union of disjoint intervals...
– zoli
Jul 19, 2018 at 7:42
• Yes, but this Lemma is a part of the proof that Lebesgue measure satisfies subaditivity; so what I’m trying to verify is essentially the statement “intervals and interval lengths behave like a measure.” Jul 19, 2018 at 9:10
• @MoedPolBollo : Now, I understand the source of misunderstanding. Just like I, nobody understood why you wanted to prove the statement at stake. I edited my answer.
– zoli
Jul 20, 2018 at 9:37