right continuous implies countably additive Let $F:\mathbb{R} \rightarrow \mathbb{R} $ be a monotonically increasing function. Let $\mu_F$ be defined on the half open real intervals such that $\mu_F((a,b]) = F(b) - F(a)$. Then $\mu_F$ is countably additive if $F$ is right-continuous. (defined here (screenshot), definition has a bigger domain but I think restricting it here still works).

But why is this? I've proved finite additivity. Also tried to use the continuity of probability measures, but I couldn't get that proof to work here since the half open intervals isn't a $\sigma$-algebra.
Here's a proof idea which I got working in the end, use the proof for countable additivity of the length function for intervals (http://www.maths.tcd.ie/~richardt/MA2224/MA2224-ch2.pdf Proposition 2.1.12), and where it uses $\epsilon/2^n$, we take a $\delta_i$ s.t. $0 &lt; t - d_i &lt;= \delta_i$ we have that $F(t) - F(d_i) &lt; \epsilon/2^i$. Nothing else changes much.

The key is the Heine Borel theorem, it reduces the countable case to a finite case, which is easy to handle.
Here's a proof idea which I got working in the end, use the proof for countable additivity of the length function for intervals (http://www.maths.tcd.ie/~richardt/MA2224/MA2224-ch2.pdf Proposition 2.1.12), and where it uses $\epsilon/2^n$, we take a $\delta_i$ s.t. $0 &lt; t - d_i &lt;= \delta_i$ we have that $F(t) - F(d_i) &lt; \epsilon/2^i$. Nothing else changes much.

The key is the Heine Borel theorem, it reduces the countable case to a finite case, which is easy to handle.