If $\mu$ is a measure, is $t\mapsto\mu([0,t])$ right-continuous with left-limits?

Let $\mu$ be a measure on $\mathcal B([0,\infty))$ with $$f(t):=\mu([0,t])<\infty\;\;\;\text{for all }t\ge0.\tag1$$

Is $f$ right- and/or left-continuous? Or does it at least have right- and/or left-limits?

We clearly want to use that a measure is continuous from below and from above: If $(A_n)$ is

1. an increasing sequence of measurable sets and $A=\bigcup_nA_n$; or
2. a decreasing sequence of measurable sets with $\mu(A_n)<\infty$ and $A=\bigcap_nA_n$,

then $\mu(A_n)\to\mu(A)$.

In order to show that $f$ is right-continuous, we would need to show that if $(t_n)_{n\in\mathbb N}\subseteq[0,\infty)$ is decreasing and $t\ge0$ with $t_n\rightarrow{n\to\infty}t$, then $f(t_n)\to f(t)$. Now, if I'm not missing something, we should have $\bigcap_n[0,t_n]=[0,t]$ and hence this should hold.

I don't think that $f$ is left-continuous, but the left limits should at least exist. Maybe we can show that if $(t_n)_{n\in\mathbb N}\subseteq[0,\infty)$ is increasing and $t\ge0$ with $t_n\to t$, then $\bigcup_{[0,t_n)}=[0,t]$?

If $t_n\nearrow t$ as $n\to\infty$, then $f(t_n)$ is a non-decreasing sequence bounded from above (by $\mu([0,t])$). Thus, $f(t_n)$ has a limit. In particular, $\lim_{n\to\infty}f(t_n)=\sup_{n\ge 1}f(t_n)$.