Here's what I am trying to prove:
Let $X$ be a topological space and $\mu$ be a finite signed Borel measure on $X$. Suppose that \begin{align*} \int f d\mu =0 \end{align*} for every continuous function $f$. Then $\mu$ is the zero measure.
Here's my take on it: If $\mu$ is an unsigned measure then then we have that $\int \chi_{X} d\mu = 0$ where $\chi_X$ is the characteristic function of $X$. Hence, we have that $\mu(X)=0$ and that forces that $\mu=0$. However, I do not see how this is true for a finite signed measure.
By Hahn Decomposition Theorem, there exists positive set $P$ and a negative set $N$ such that $X$ is the disjoint union of $P$ and $N$. However, I cannot conclude that $\chi_P$ and $\chi_N$ are continuous to do the same as in the case of unsigned measure.
Hints are appreciated!