 Solution

A distribution function is nondecreasing and right-continuous. Every nondecreasing function has finite left limits, so $F$ is càdlàg. Along each partition its increments are nonnegative and telescope, givingFor each $n$, the set $A^{(n)}=\{s\in [0,n]:\Delta F(s)\ge 2^{-n}\}$ is finite because the sum of its positive jumps is at most $F(n)-F(0)$. Every jump belongs to some $A^{(n)}$, so $A_F={\bigcup }_n{A}^{(n)}$ is countable.

For a finite partition, the identity $y^2-x^2=2y(y-x)-(y-x{)}^2$ givesThe first sum tends to the Lebesgue-Stieltjes integral $2{\int }_0^{t}FdF$. In the second, intervals containing no prescribed large jump contribute at most their largest increment times $F(t)-F(0)$; first retain finitely many jumps above a threshold and then let the threshold vanish. The limit is therefore ${\sum }_{s\in A_F\cap (0,t]}|\Delta F(s){|}^2$, proving