Given a measure $\mu$ that satisfies $\mu((a,b]) = F(b) - F(a)$ for a specified increasing, right-continuous function $F$, find $\mu(\mathbb{R}$.

Question

Define $F : \mathbb{R} \to \mathbb{R}$ by

$$F(x) : = \begin{cases} 0 & \quad x < 1 \\ \frac{2x}{1+x} & \quad x \ge 1.\end{cases}$$

Then $F$ is increasing and right continuous. Suppose that $\mu$ is a measure that satisfies $\mu((a,b]) = F(b) - F(a)$, where $a<b$. Find $\mu(\mathbb{R}$).

My attempt:

Write $\mathbb{R} = \bigcup_{n=1}^\infty{(-n,1-n]} \cup \bigcup_{n=1}^\infty{(n-1,n]}$. Then $\mathbb{R}$ is a countable union of pairwise disjoint measurable sets. Hence, \begin{align*}\mu(\mathbb{R}) & = \mu\left(\bigcup_{n=1}^\infty{(-n,1-n]} \cup \bigcup_{n=1}^\infty{(n-1,n]}\right)\\ & = \mu\left(\bigcup_{n=1}^\infty{(-n,1-n]}\right) + \mu\left({\bigcup_{n=1}^\infty{(n-1,n]}}\right)\\ & = \sum_{n=1}^\infty{{F(1-n) - F(-n)}} + \sum_{n=1}^\infty{{F(n) - F(n-1)}}\\ & = 0 + \sum_{n=1}^\infty{F(n)}-\sum_{n=1}^\infty{F(n-1)}\\ & = \sum_{n=1}^\infty{F(n)}-\sum_{n=1}^\infty{F(n)}\\ & = 0. \end{align*}

However, I am not sure if I did this correctly.

Answer 1

You are correct in that $$ \mu(\mathbb R) = \sum_{n=1}^\infty F(n) - F(n-1). $$ From here, we have \begin{align} \sum_{n=1}^\infty F(n) - F(n-1) &= \lim_{n\to\infty}\sum_{k=1}^n F(k)-F(k-1)\\ &= \lim_{n\to\infty} F(n) - F(0)\\ &= 2. \end{align}