Lebesgue-Stieltjes measure function is a bijection

Question

I came across this exercise in my Measure Theory workbook and I've been stuck on it. This is the question :

Let F be the set of all non-decreasing right-continuous functions $f : \mathbb{ R} \rightarrow \mathbb{R}$ with $f(0) = 0$. Let $\mathcal M$ be the set of all measures $\mu$ on $(\mathbb R, \mathcal B)$ that are finite on every bounded interval (that is, $\mu( [a, b] ) < \infty$ for any $a,b \in \mathbb R$ with $a < b$).

Show that the map $\Lambda : \mathcal F \rightarrow \mathcal M$, which sends $f \in \mathcal F$ to its Lebesgue-Stieltjes measure $\lambda_f \in \mathcal M$, is a bijection and describe how to construct its two-sided inverse $F : \mathcal M \rightarrow \mathcal F$.

(You can assume without proof that, for every $f \in \mathcal F$ and reals $a < b$, we have $\lambda_f \in \mathcal M$ and $\lambda_f ( (a, b] ) = f(b) − f(a)$.)

Thanks

Answer 1

I am also trying to show the same problem. It is straightforward to show injectivity:

firstly take two functions $f_1, f_2$ that are not equal, that is to say they differ at some point $x\in{\mathbb{R}}$ s.t. $f_1(x)\neq{f_2(x})$. Now consider the interval $(0,x]$. Observe how $\lambda_{f_1}((0,x])=f_1(x)-f_1(0)=f_1(x)-0=f_1(x)$. Similarly $\lambda_{f_2}((0,x])=f_2(x)-f_2(0)=f_2(x)-0=f_2(x)$. And as these functions differ at one point $x$. $\lambda_{f_1}\neq{\lambda_{f_2}}$.

I am not sure how to show surjectivity though.

Answer 2

for surjectivity I think try:

Let $\mu \in M$ then set $f(x)= \mu ((0,x])$ for positive $x$ similar for negative. then show f must be increasing and right cts.