this is my problem. Most of books take LS measure as done or at least take lost of round to prove it with Caratheodory things. Text provided by my university has left as excersice the proof of the measure property of the LS formula.

Text says: Let $g:\mathbb{R} \rightarrow \mathbb{R}$ be a increasing function.

Define for each interval $[a,b)$ of $\mathbb{R}$, with $a,b \in \mathbb{R}, a<b$, the value of the function $\mu$ as follow: $\mu([a,b))= g(b)-g(a)$, also define $\mu(\emptyset)=0$

Now, we define $\rho:\mathcal{P}(\mathbb{R})\rightarrow[0,\infty]$ as follow:

$$\rho(E)= \inf \left\{\sum\limits_{n=1}^{\infty}\mu([a_n,b_n)):E\subseteq\bigcup\limits_{n=1}^{\infty}[a_n,b_n) \right\}$$

I must show that $\rho$ is in fact an outrer measure in $\mathbb{R}$.

I was attempting to make a kind of "telescopic sum" to find out the measure of $\emptyset$ is zero but my brain is fried. Also tried to make it equal to another exterior sum but failed on the inequalities.

Any hint, help or text is very welcome.

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    $\begingroup$ The website always cuts my "Hi there", sorry if it looks so rude. $\endgroup$ Commented Apr 24, 2020 at 8:00
  • 1
    $\begingroup$ (I'm pretty sure that cutting the "Hi there" is a deliberate design choice, don't need to worry about that) $\endgroup$ Commented Apr 24, 2020 at 8:03
  • $\begingroup$ can't you choose $a_n=b_n$ for every $n$ to get a sequence of intervals for which the sum in the infimum is $0$ already? to prove $\rho(\emptyset ) =0$ I mean. $\endgroup$
    – Targon
    Commented Apr 24, 2020 at 8:09
  • $\begingroup$ No, because they ask a and b to be a < b wich includes $a_n$ and $b_n$ $\endgroup$ Commented Apr 24, 2020 at 8:12
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    $\begingroup$ @JorgeArturoQuiroz Regarding removing "Hi there", it's not because it's considered to be rude. You may wish to read What's wrong with using polite language in questions? and, in particular, this answer of the post it's closed as being a duplicate of. $\endgroup$ Commented Apr 24, 2020 at 8:13

2 Answers 2


To show that in fact $\rho (\emptyset) = 0$ you can basically apply epsilon's answer. To show monotonicity, i.e. $E \subset F \Rightarrow \rho (E) \leq \rho (F)$, observe that whenever $E \subset F$ and $F \subset \cup_n [a_n , b_n)$ we also have $E \subset \cup_n [a_n , b_n )$ and therefore $\rho(E) \le \rho(F)$.
Now to show that $\rho$ is countably sub-additive let $A_1,A_2,\ldots \in \mathcal P (\mathbb R )$ be measurable sets. Let $\big( I_k^n \big)_{k\in \mathbb N}$ be a sequence of intervals (left closed, right open) that cover $A_n$. Then we have $$\bigcup_{n \in \mathbb N} A_n \subset \bigcup_{n,k\in \mathbb N} I_k^n .$$Therefore, given $\varepsilon >0$, if we choose above $\big(I_k^n\big)_k$ such that $\rho (A_n) > \sum_k \mu(I_k^n) - \varepsilon2^{-n}$ we arrive at $$\rho \left( \bigcup_{n \in \mathbb N } A_n \right) \leq \sum_n \sum_k \mu \big( I_k^n \big) \leq \sum_n \rho (A_n )+ \varepsilon 2 ^{-n} = \varepsilon + \sum_{n\in \mathbb N} \rho (A_n).$$


A general fact about monotone functions on $\mathbb R$ is that they only have countably many discontinuities. Thus, there exist (uncountably many) points where $g$ is continuous.

Let $x_0$ be a point where $g$ is continuous. For every $\epsilon > 0$ we may choose $x_0 < x_\epsilon$ such that $\mu ([x_0, x_\epsilon)) = g(x_\epsilon) - g(x) < \epsilon$

Thus, $\rho(\varnothing) < \epsilon$ for every $\epsilon > 0$.


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