Consider the measure space $\big(\mathbb{R_+},\mathcal{B}(\mathbb{R_+}),\mu\big)$, $\; b=\mu (\mathbb{R_+})$, possibly $b=+\infty$. Also assume $\mu\big([0,t]\big)<\infty \; \; \forall t \in \mathbb{R_+}$.
Define
\begin{equation} c(t)=\mu\big([0,t]\big), \;\; t \in \mathbb{R_+} \end{equation}
Provided that $c$ is non-decreasing, show that $c$ is right-continuous, i.e., show that for any sequence $(t_n)_{n=1}^{\infty} \subset \mathbb{R_+}$ such that $t_n \searrow t \in \mathbb{R_+}$, it holds \begin{equation} c(t_n) \searrow c(t) \end{equation}
This question is taken from exercise I.5.14 of "Probability and Stochastics" by E. Cinlar.
My attempt
I would rely on the sequential continuity of measures, but it only holds for increasing sequences of sets if the measure $\mu$ is not finite. Precisely, I would like to write as follows:
Fix $t \in \mathbb{R_+}$ and consider any sequence $(t_n)_{n=1}^{\infty} \subset \mathbb{R_+}$ decreasing to $t$. Then
\begin{equation} t_n \searrow t \implies [0,t_n] \searrow [0,t] \implies \mu\big([0,t_n]\big) \searrow \mu\big([0,t]\big) \implies c(t_n) \searrow c(t) \end{equation}
But in fact sequential continuity of measures is defined for increasing sequences of sets, and it can be applied to decreasing sequences of sets only when $\mu\big(\mathbb{R}\big) = b < \infty$, so that
\begin{equation} \mu\big([0,t_n]\big) = b - \mu\big((t_n, \infty]\big) \searrow b - \mu\big((t,\infty]\big) = \mu\big([0,t]\big) \end{equation}
because I could apply sequential continuity of $\mu$ on the increasing sequence of sets $(t_n, \infty])$ so that
\begin{equation} \mu\big((t_n, \infty]\big) \nearrow \mu\big((t,\infty]\big) \end{equation}
So, not only I cannot apply sequential continuity to decreasing sequences in order to get right-continuity of $c$ when $\mu\big(\mathbb{R}\big)=\infty$, but I even get left-continuity of $c$ from the sequential continuity of $\mu$ over increasing sequences of sets.
Any help in proving the right-continuity of $c$ would be much appreciated, as I have been stuck on this for a while.
Let $t_0\in\mathbb R_+$ be given, and $t_n\downarrow t_0$. Since we assume $\mu([0,t])<\infty$ for every $t$, by assumption, $\mu([0,t_0+1])<\infty$. There exists $N = N(t_0)$ such that for $n>N(t_0)$, we have $t_n<t_0+1$. Therefore, without loss of generality, we can replace $\mu$ with the restricted measure $d\mu' = 1_{[0,t_0+1]}d\mu$, where $\mu'$ is now a finite measure on $\mathbb R_+$ with $\mu'([0,s]) = \mu([0,s])$ for all $s\le t_0+1$. In particular, $\mu'([0,t_n]) = \mu([0,t_n])$ for all $n > N(t_0)$.
By what's written above, it suffices to prove the claim for the finite measure $\mu'$, which I assume you know how to handle by taking complements and using continuity from below of measures. The key to this reduction is that the continuity is a "local" statement, and the measure you are given in the assumption is "locally finite."