I'm trying to prove this characterization. Could you have a check if my attempt is fine?
Let $X$ be a topological space and $\mu$ a signed Borel measure on $X$. Let $\mu = \mu_+ - \mu_-$ be the Jordan decomposition of $\mu$. Let $L = \{f:X \to [0, 1] \mid f \text{ is measurable}\}$. Then $$ \mu_+(X) = \sup_{f \in L} \int_X f \mathrm d \mu. $$
My attempt: We have $$ \begin{align} \mu_+ (X) &= \sup \{\mu(B) \mid B \in \mathcal B (X)\} \\ &= \sup_{B \in \mathcal B (X)} \int_X 1_B \mathrm d \mu \\ &\le \sup_{f \in L} \int_X f \mathrm d \mu \quad \text{because} \quad 1_B \in L. \end{align} $$
On the other hand, $$ \begin{align} \sup_{f \in L} \int_X f \mathrm d \mu &= \sup_{f \in L} \int_X f \mathrm d (\mu_+ - \mu_-) \\ &= \sup_{f \in L} \left ( \int_X f \mathrm d \mu_+ - \int_X f \mathrm d \mu_- \right )\\ &\le \sup_{f \in L} \int_X f \mathrm d \mu_+ \\ &\le \int_X \mathrm d \mu_+ \\ &= \mu_+ (X). \end{align} $$
This completes the proof.