I want to show that there exists a signed measure $\mu$ on $(\mathbb{R},\mathcal{B})$ such that $\mu((-\infty,x])=e^{3x}-e^x$ holds.
I already know that $\mu$ is defined on $\mathcal{B}$ since $\mu$ is defined on $\{(-\infty,x]|x\in\mathbb{R}\}$. Also $\mu:\mathcal{B}\rightarrow (-\infty,\infty]$. Therefore I only have to show the sigma additivity of $\mu$.
I think I have to define a specific $\mu$ with $\mu((-\infty,x])=e^{3x}-e^x$ such that I can show the sigma additivity. Can anyone help me?
Hint: Define $\mu (E)=\int_{E\cap (-\infty, 0]} f(t)dt +\int_{E\cap (0,\infty)} f(t)dt$ where $f(t)=3e^{3t}-e^{t}$. The first term gives a countably additive function because of integrability of $f$ on the negative real line. The second part is countably additive by Monotone Convergence Theorem since $f(x) > 0$ for $x >0$.