I have the following problem:
Let $\mathcal{F}$ be the algebra of finite disjoint unions of right-semiclosed intervals of $\mathbb{R}$, and define the set function $\mu$ on $\mathcal{F}$ as follows: \begin{align*} \mu(-\infty,a] &= a, &&a\in\mathbb{R}\\ \mu(a,b] &= b-a, &&a,b\in\mathbb{R}, a<b\\ \mu(b,\infty) &= -b, &&b\in\mathbb{R}\\ \mu(\mathbb{R}) &= 0 \end{align*} (a) Show that $\mu$ is finitely additive but not countably additive on $\mathcal{F}$.
(b) Show that $\mu$ is finite but unbounded on $\mathcal{F}$.
The letter (a) is quite straightforward. I just need to show that $$\mu\left(\bigcup\limits_{i=1}^{n} A_i\right)=\sum_{i=1}^n\mu(A_i)$$ And that it's not valid for $n\to\infty$
For the letter (b), that $\mu$ is finite it's ok, but I don't know how to show it's unbounded. Anyone could help me?
To show that $\mu$ is unbounded, we need to show that for any $M > 0$, there exists some interval $(a,b]$ such that $\mu(a,b] > M$. So, fix $M > 0$ and take $$ a < -\frac{M}{2} \qquad\text{and}\qquad b > \frac{M}{2}. $$ Then we have $$ \mu(a,b] = b-a > \frac{M}{2} + \frac{M}{2} = M, $$ which gets the job done.