Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 202 / 1

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

Refine a dyadic partition by inserting $s$ and $t$. The triangle inequality shows that its variation over $[s,t]$ is at least $|f(t)-f(s)|$. Passing to the defining limit givesConsequently the càdlàg functionsare nondecreasing: for $s\le t$, the displayed inequality makes both increments nonnegative. Thus they are distribution functions in the Stieltjes sense and $f=F-G$; this is the Jordan decomposition of a function of bounded variation.

Now $A_f\subseteq A_F\cup A_G$, so $A_f$ is countable by part (a). For any finite subset $J\subseteq A_f\cap (0,t]$, partitions isolating its points and the triangle inequality giveTaking the supremum over finite $J$ proves ${\sum }_{s\in A_f\cap (0,t]}|\Delta f(s)|<\infty$.

Apply part (a)'s square formula, extended by the Jordan decomposition of a function of bounded variation from nondecreasing functions to arbitrary càdlàg functions of bounded variation, to $f+g$, $f$, and $g$. Sincelinearity of the Lebesgue-Stieltjes integral leaves ${\int }_0^{t}fdg+{\int }_0^{t}gdf$. At each time $s$, polarization of the jump correction givesOnly common jump times contribute, and hence