I’m going through the construction of Lebesgue measure, which relies on this result. In the following, the $I$ are all intervals defined on $[0,1]$ and $\mathbb{P}(I)$ denotes the length of the interval.

Lemma 1

Let $I_1, I_2, \cdots, I_n$ be a finite collection of intervals whose union contain an interval $I$, then $$\sum_{j=1}^{n} \mathbb{P}(I_j) \geq \mathbb{P}(I).$$


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 (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, in the sum of interval lengths:

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

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

Lemma 2:

Let $I_1, I_2,\cdots$ be a countable collection of open intervals, whos union contains a closed interval, then $$\sum_{j\geq 1}\mathbb{P}(I_j) \geq \mathbb{P}(I).$$


By the Heine-Borel theorem, for any countable collection of open sets $\{O_n\}$ and some closed set $ \bigcup_{n\geq1} O_n \supseteq O$, there exists a finite sub cover $$O_{a_1} \cup O_{a_2} \cup \cdots \cup O_{a_k} \supseteq O.$$

Thus $$ \bigcup_{j=1}^\infty I_j \supseteq I. \implies \bigcup_{a=1}^k I_{j_a} \supseteq I$$ and the result follows from lemma 1.


Lemma 3:

For any countable collection of intervals $\{I_j\}$, whose union contains an interval $I$, $$\sum_{j \geq 1} \mathbb{P} (I_j) \geq \mathbb{P}(I).$$


Choose some $\epsilon >0$ and extend the left and right end points of the $I_j$ by $\epsilon 2^{-j}$ to form an open set $$I_{j_\epsilon}=\left(a_j -\epsilon 2^{-j} , b_j + \epsilon 2^{-j} \right).$$

Next, extend the end points of $I$ by $\epsilon$ to make the closed set $$I_\epsilon =[a- \epsilon , b + \epsilon].$$

Since $\bigcup_{j \geq 1} I_j \supseteq I,$ there exists some $a_k$ and $b_l$ such that

$$a_k <a \leq b < b_l.$$

Further, since $k$ and $l$ are both $\geq 1$, we can choose $\epsilon$ small enough that
$$a_k - \epsilon 2^{-k}< a_k<a - \epsilon \quad, \quad b+\epsilon < b_l < b_l + \epsilon 2^{-l}.$$

Thus, we have constructed a countable collection of open intervals $\{I_{j_\epsilon}\}$, whose union contains the closed interval $I_\epsilon$. By lemma 2:

$$\sum_{j\geq1} \mathbb{P}(I_{j_\epsilon}) \geq \mathbb{P}(I_\epsilon)$$

$$\implies \sum_{j \geq 1} \Big \{ \big( b_j + \epsilon 2^{-j} \big) - \big(a_j - \epsilon 2^{-j} \big)\Big\} \geq( b + \epsilon )- (a - \epsilon).$$

$$\implies \sum_{j \geq 1} ( b_j - a_j) + \epsilon \sum_{j \geq 1} \big (2^{1-j} \big) \geq (b-a) + 2\epsilon.$$

$$ \implies \sum_{j \geq 1} ( b_j - a_j) + 2\epsilon \geq (b-a) + 2\epsilon.$$

$$\implies \sum_{j\geq 1} \mathbb{P}(I_j)=\sum_{j \geq 1} ( b_j - a_j) \geq (b-a) = \mathbb{P}(I).$$ $ \square$

  • $\begingroup$ What's the question? $\endgroup$ Jul 17, 2018 at 5:37
  • $\begingroup$ @LordSharktheUnknown I just want to make sure the proof is correct, as it is necessary for the construction of Lebesgue measure on $[0,1]$. $\endgroup$
    – Mo Pol Bol
    Jul 17, 2018 at 5:39
  • $\begingroup$ What does "Choose $a_k$ and $b_k$ such that: $a_k\le a_j$, $b_j\le b_k$" mean? $\endgroup$ Jul 17, 2018 at 5:42
  • $\begingroup$ @LordSharktheUnknown That’s a poorly worded way of saying $a_k$ is the smallest element of the set of left end points and $b_k$ is the largest element of the set of right end points. $\endgroup$
    – Mo Pol Bol
    Jul 17, 2018 at 5:45
  • $\begingroup$ Then that's an error straight away; the least left endpoint may not belong to the same interval $I_k$ as the largest right endpoint. $\endgroup$ Jul 17, 2018 at 5:46

1 Answer 1


Prove Lemma 2 directly, using a finite recursion, and you will not need Lemma 1. As follows: Let $I=[a,b]$ with $a<b$ and let $C=\{c_j:1\leq j\leq n\}$ be a family of open intervals that covers $[a,b].$

Let $f(1)$ be the least $m$ such that $0\in c_m.$

For $j\geq 1,$ if $\sup c_{f(j)}\leq b,$ let $f(j+1)$ be the least $m$ such that $\sup c_{f(j)}\in c_m.$ If $\sup c_{f(j)}>b$ then $f(j')$ is not defined for $j'>j$.

By induction on $j:$ If $f(j)$ exists then

$(\alpha).\;\sup (\cup_{i\leq j}c_{f(i)})=\sup c_{f(j)},$ and

$(\beta).\; \cup_{i\leq j}c_{f(i)}\supset [0,\sup c_{f(j)}), $ and

$(\gamma).\;$ If $1\leq i<j$ then $f(j)\ne f(i) $ and $c_{f(i)}\ne c_{f(j)}$.

Now dom($f)=\{1,..,K\}$ for some $K\leq n.$ And $\sup c_{f(K)}>b.$ For convenience define $U(0)=a$ and $U(j)=\sup c_{f(j)}$ for $1\leq j\leq K.$

For $1\leq j\leq K $ we have $U(j-1)\in c_{f(j)}$ so we have $P(c_{f(j)})>(\sup c_{f(j)})-U(j-1)=U(j)-U(j-1).$

Therefore, (especially by $(\gamma)\;$), we have $\sum_{i=1}^nP(c_i)\geq$ $ \sum_{j=1}^KP(c_{f(j)})>$ $\sum_{j=1}^K U(j)-U(j-1)=$ $=U(K)-U(0)=U(K)-a>b-a=P(I).$

  • $\begingroup$ To the proposer: You seem to be having difficulties with Lemma 1. But your main result (Lemma 3) uses Lemma 2 only, so I decided to post a direct proof of Lemma 2 . $\endgroup$ Jul 20, 2018 at 12:06
  • $\begingroup$ Thanks for the alternative proof! I am still trying to understand all of it. A few questions for clarification: (1) Is your definition of $C$ not that of a finite collection of open intervals? (2) By “let $f(j+1)$ be the least $m$ such that $\sup c_{f(j)} \in c_m$” are you essentially specifying an overlap of intervals similar to that in my proof of Lemma 1? Thanks again. $\endgroup$
    – Mo Pol Bol
    Jul 21, 2018 at 7:52

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