Problem: Let $F(x)$ be a continuous and non-decreasing function on $[0,1]$ with $F(0)=0$ and $F(1)=1$. I'd like to prove the existence of a couuntably additive measure $\mu$ on Borel subsets of $[0,1]$ s.t. $\mu\left([a,b]\right)=F(b)-F(a)$ for all intervals $[a,b]\subset[0,1]$.

I can prove the case when $F(x)$ is the distribution function, which is non-decreasing and right continuous on the real line. Then, $F(x)=\mu\left((-\infty,x]\right)$. $\mu\left((a,b]\right)=F(b)-F(a)$.

My sketch of the proof for this case: We need to show that if $(a,b]=\bigcup_{j=1}^{\infty}(a_j,b_j]$, then $F(b)-F(a)=\sum_{j}F(b_j)-F(a_j)$.

$\ge$ is obvious.

For $\leq$, we apply Caratheodory extension for $\mu$ from the semiring of intervals to the Borel $\sigma$-field. By definition of distibution function, $F(x)\to 0$ as $x\to-\infty$. We can replace $a$ by finite number of $a'$ s.t. $F(a')-F(a)<\varepsilon$. Similary, we can replace $b$ by finite $b'$ s.t. $F(b)-F(b')<\varepsilon$. By right continuity of $F(x)$, we can replace $(a_j,b_j]$ by $(a_j,b_j')$ s.t. $F(b_j')-F(b_j)<\frac{\varepsilon}{2^j}$. Now we have $$F(b')-F(a')\ge F(b)-F(a)-2\varepsilon$$ and $[a',b']\subset(a,b]$ is closed and bounded.

Moreover, $(a_j,b_j')$ is an open cover of $[a',b']$. By Heine-Borel, we can find such finite subcover and $$\sum_{j}F(b_j)-F(a_j)\ge\sum_{j}F(b_j')-F(a_j)-\frac{\varepsilon}{2^j}\ge F(b')-F(a')-2\varepsilon\ge F(b)-F(a)-3\varepsilon$$ Let $\varepsilon\to 0$ to get the desired result.

My questions are:

  1. How can I adapt the proof I provided above for the problem I am asking? I am stuck, since countable union of disjoint closed sets may not be a closed set. And the problem seems to be a restriction of the distribution function to the interval $[0,1]$. If so, how? And if not, how can I write a new proof?

  2. I am also interested in the case when $F(x)$ has jumps on its domain $[0,1]$. Do I have break into the case on finite and infinite jumps? I know the fact that Dirac measures are countably additive, so does this fact help to prove this case?

Added(09/13/2020): I think it is possible for singular points to have non-zero measures. For example in my case, we can define Dirac measure at $x=\frac{1}{2}$, a valid countably additive measure on $\mathcal{B}$. $$F(x)=\chi_{\{[1/2,1]\}}$$ is one such example. Please point out if I have any flaw in my reasoning.

If there is a similar question regarding this, please direct me to that. I know this may be a basic question, but I'd like to receive help on this. Thank you.

  • $\begingroup$ You can't. Take $$F(x) = \begin{cases} 0 &\text{if } x < 1/2, \\ \frac{1}{2} &\text{if } x = 1/2, \\ 1 &\text{if } x > 1/2. \end{cases}$$ Then various properties of measures force a) $\mu(\{1/2\}) = 0$, b) $\mu(\{1/2\}) = 1/2$, and c) $\mu(\{1/2\}) = 1$. Some additional conditions or some change in requirements is necessary. $\endgroup$ Sep 8, 2020 at 15:23
  • $\begingroup$ Sorry, I overlooked the continuity requirement. With that, $F$ is a distribution function, and the usual way works. If you don't know how to deal with distribution functions defined only on an interval, extend it to $\mathbb{R}$ by setting $F(x) = 0$ for $x < 0$ and $F(x) = 1$ for $x > 1$. $\endgroup$ Sep 8, 2020 at 15:50
  • $\begingroup$ @DanielFischer, Hello, can you write an answer regarding my two question? I don't know how to deal with the closed set $[a,b]$, since distribution function is defined for half-open half-closed intervals. Thank you! $\endgroup$
    – Mike
    Sep 8, 2020 at 15:53
  • $\begingroup$ I don't understand your first question. If you have $\mu((a,b]) = F(b) - F(a)$ you have $\mu([a,b]) = F(b) - F(a)$ by continuity. Where does a countable union of disjoint closed sets enter? $\endgroup$ Sep 8, 2020 at 15:58
  • $\begingroup$ @DanielFischer: Hello, my confusion is how to possibly link my prove for $\mu\left((a,b]\right)=F(b)-F(a)$ to my problem $\mu\left([a,b]\right)=F(b)-F(a)$? And does such countably additive measure exists if $F(x)$ admits jumps in its domain $[0,1]$? Could you possibly write an answer regarding my confusion. That'll clear things up. $\endgroup$
    – Mike
    Sep 8, 2020 at 16:06

1 Answer 1


Okay, I guess I kind of figure it out. Please let me know if there is any mistake. Basically, you do all the measure theory things over $\mathbb{R}$ firstly, just like the first part your post.

Then, you get a unique Borel measure $\lambda$ on the whole real line $\mathbb{R}$, and the definition of it is $$\lambda((a,b])=F(b)-F(a).$$

Now, restrict $\lambda$ to $[0,1]$. That is, you define $\mu:=\lambda|_{[0,1]}$. But this is not over, you still need to argue why $$\mu([a,b])=F(b)-F(a),$$ since currently the definition is $\mu((a,b])=F(b)-F(a).$

This involves the continuity. As pointed out by you, if the function has jump, you can have a singular measure at a point. This involves the following lemma.

Lemma: Let $F$ be an increasing and right-continuous function, and $\mu$ to be the measure associated to it. Then, $\mu(\{a\})=F(a)-F(a-)$, $\mu([a,b))=F(b-)-F(a-)$ and $\mu([a,b])=F(b)-F(a-).$

You could find the proof here http://www.math.ttu.edu/~drager/Classes/01Fall/reals/ans2.pdf.

However, if you read the proof, you will see that by definition of $F(a-)$, if your function is continuous, then $F(a-)=F(a)$ and therefore, for continuous function, you can arrive conclusion $$\mu([a,b])=F(b)-F(a-)=F(b)-F(a).$$ However, for jump-function, if it is right continuous, then you can only get the lemma, which is a little bit weaker.

For left continuous function, your construction of Borel measure over $\mathbb{R}$ will use $[a,b)$, so the direction will change. You could easily modify the statement and the proof to discuss the left-continuous case, but the idea is the same so I will not bother to write it here.


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