Let $\mu \neq 0$ be a signed, finite Borel measure on $[0,1]$ that is not concentrated at $0$.
Does there exists $a \in (0,1)$ such that $\mu([a,1]) \neq 0$?
If there is no such $a$, then $\mu((b,c)) = 0$ for all $b,c \in (0,1)$, hence $\mu(A) = 0$ for all Borel sets in $[0,1]$, but does this imply that $\mu = 0$?
Suppose there is no such $a$. Then $\mu [a,b)=0$ whenever $a< b <1$ which implies $\mu (A)=0$ for every Borel set $A \subseteq (0,1)$. [Finite disjoint unions of half closed intervals $[a,b)$ form an algebra which generates the Borel sigma algebra. By Uniqueness part of Cartheodory Extension Theorem we get $\mu (A)=0$ for every Borel set $A $ in $(0,1)$]. Also $\mu \{1\}=\lim \mu (1-\frac 1 n ,1]=0$. So $\mu$ is concentrated on $\{0\}$ which is a contradiction .