If $\mu$ is a finite Borel measure on $\mathbb{R}$, then the function $F^{\mu}: \mathbb{R} \rightarrow \mathbb{R}$ , $F^{\mu}(\lambda) = \mu((-\infty, \lambda])$ is right-continuous.
Attempt at solution
I need to show that $\lim \limits_{\lambda \to c^{+}} F^{\mu}(\lambda) = F^{\mu}(c)$. Using the $(\epsilon, \delta)$ formulation of limits, I consider the inequality $$F^{\mu}(\lambda) - F^{\mu}(c) < \epsilon$$ $$\mu((-\infty, \lambda]) - \mu((-\infty, c]) < \epsilon$$ $$\mu((c, \lambda]) < \epsilon$$
where I obtained the last step using $\mu(A) = \mu(B) + \mu(A\setminus B)$, which in turn derives from the union-sum propriety of Borel measures for disjoint sets.
But now I seem to be stuck due to the generality of $\mu$: if it were the usual measure $\mu((a,b]) = b-a$, then the result would immediately follow with $\delta = \epsilon$, but since here I don't have an explicit expression for $\mu$ I don't know how to get to an inequality involving the argument of $\mu$, which is what one usually does in similar situations. What general propriety of Borel measures am I missing which allows one to arrive at the result?
It follows from continuity of measure, that is if $A_n \downarrow A$ (with $\mu A_1 $ finite, which is the case here) we have $\mu A_n \to \mu A$.
Let $\lambda_n \downarrow \lambda$, and $A_n = (-\infty, \lambda_n]$, then $A=\cap_n A_n = (-\infty, \lambda]$ and so $F^\mu(\lambda_n) \downarrow F^\mu(\lambda)$.