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

Mathematics Asked on February 11, 2021

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. $muleft([a,b]right)=F(b)-F(a)$ 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, $F(x)=muleft((-infty,x]right)$. $muleft((a,b]right)=F(b)-F(a)$.

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 $xto-infty$. 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
$$F(b’)-F(a’)ge F(b)-F(a)-2varepsilon$$ 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
$$sum_{j}F(b_j)-F(a_j)gesum_{j}F(b_j’)-F(a_j)-frac{varepsilon}{2^j}ge F(b’)-F(a’)-2varepsilonge F(b)-F(a)-3varepsilon$$
Let $varepsilonto 0$ 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.

One Answer

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.

Answered by JacobsonRadical on February 11, 2021

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