Suppose that for every $T > 0$ we have a finite signed measure $\mu_T : \mathcal{B}([0, T]) \to (-\infty, \infty)$ and that for all $T_1 < T_2$ we have that $\mu_{T_2}(A) = \mu_{T_1}(A)$ for all $A \in \mathcal{B}([0, T_1])$. Can we show that there exists a unique (probably $\sigma$-finite) signed measure $\mu : \mathcal{B}([0, \infty)) \to [-\infty, \infty]$ such that the restriction to $\mathcal{B}([0, T])$ for each $T > 0$ is equal to $\mu_T$?
My first thought would be to use Carathéodory's extension theorem on the positive and real parts of $\mu_T$, however $\mathcal{B}([0, T])$ is not a ring on $[0, \infty)$, even if we add the complements and extend it properly, I'm not sure if it is a ring either.
Any help is appreciated.
I don't think so. Let $\lambda$ be the standard Borel measure and for $T\geq 0$ consider $$\mu_T(A)=\lambda(A\cap [0,1))-\lambda(A\cap [1,2))+\lambda(A\cap [2,3))-\lambda(A\cap [3,4))+\dots +(-1)^{\lfloor T \rfloor}\lambda(A\cap [\lfloor T \rfloor, T]).$$ Then $\{\mu_T\}_T$ satisfies your consistency conditions, but if $\mu$ is any extension we would have $\mu(\bigcup\limits_{n\in \mathbb{N}} [2n, 2n+1))=\infty$ and $\mu(\bigcup\limits_{n\in \mathbb{N}} [2n+1, 2n+2))=-\infty$. This is a contradiction since a signed measure cannot attain both of the values $\infty$ and $-\infty$ at the same time.