# existence of countably additive measure for Borel subsets of $[0,1]$

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.

• 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. Sep 8, 2020 at 15:23
• 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 $x > 1$. Sep 8, 2020 at 15:50
• @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!
– Mike
Sep 8, 2020 at 15:53
• 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? Sep 8, 2020 at 15:58
• @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.
– Mike
Sep 8, 2020 at 16:06

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.