A nondecreasing function splits uniquely into a jump part and a continuous part

If $a=b$, the definition gives $J_F(a)=0$ and $C_F(a)=F(a)$, so all five claims are immediate on the singleton interval. Hence assume $a<b$ from now on.

If $a<x<y\le b$, every finite pair $(S,T)$ with $S\subseteq (a,x]$ and $T\subseteq (a,x)$ is also admissible for $y$, so $J_F(x)\le J_F(y)$. The comparison with $x=a$ follows from $J_F(a)=0$ and the nonnegative definition of $J_F(y)$. Thus $J_F$ is nondecreasing on $[a,b]$.

Fix $a\le x<y\le b$. First suppose $x>a$, and split any finite pair $(S,T)$ admissible in the definition of $J_F(y)$ into its old part $S_0\subseteq (a,x]$, $T_0\subseteq (a,x)$ and its new part $S_1\subseteq (x,y]$, $T_1\subseteq [x,y)$. The old contribution, including ${\beta }_a$, is at most $J_F(x)$. Order the distinct points of $S_1\cup T_1$. Monotonicity of $F$ makes the new jump contributions telescope through disjoint successive value intervals from $F(x)$ to $F(y)$; this includes the possible right jump $F(x^{+})-F(x)$ when $x\in T_1$, and gives a total at most $F(y)-F(x)$. Taking the supremum over $(S,T)$ yields $J_F(y)-J_F(x)\le F(y)-F(x)$. If $x=a$, the endpoint contribution ${\beta }_a=F(a^{+})-F(a)$ followed by the jumps in $(a,y]$ and $(a,y)$ telescopes in the same way, giving more precisely $0\le J_F(y)-{\beta }_a\le F(y)-F(a^{+})$ and hence $J_F(y)\le F(y)-F(a)$. Therefore the increment inequality holds for every $a\le x<y\le b$, and $C_F(y)-C_F(x)=F(y)-F(x)-(J_F(y)-J_F(x))\ge 0.$ Thus $C_F$ is nondecreasing, proving claim 1.

By Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used, the discontinuity set of $F$ in $(a,b)$ is at most countable; index it without repetitions as $(s_n{)}_{n\in A}$ for some $A\subseteq \mathbb{N}$. At points outside that set both interior one-sided jump sizes vanish. The only remaining possible contribution in the defining supremum is the left jump at $b$, which occurs exactly when $x=b$. Thus The jump function of a nondecreasing function on a compact interval agrees with the nonnegative series $J_F(x)={\beta }_a+{\sum }_{n\in A,s_n\le x}(F(s_n)-F(s_n^{-}))+{\sum }_{n\in A,s_n<x}(F(s_n^{+})-F(s_n))+{\mathbf{1}}_{\{b\}}(x)(F(b)-F(b^{-}))(x>a)\mathrm{.}$ This is claim 2.

Fix $c\in (a,b)$. Step 3.1 shows that $J_F(c)-J_F(c^{-})=F(c)-F(c^{-}),J_F(c^{+})-J_F(c)=F(c^{+})-F(c)\mathrm{.}$ Therefore $C_F(c)-C_F(c^{-})=0,C_F(c^{+})-C_F(c)=0.$ Since $C_F$ is nondecreasing, One-sided limits of a monotone function always exist: for $f$ nondecreasing on an interval $I$ and $c\in I$, ${\lim }_{x\to c^{-}}f(x)=\sup \{f(x):x\in I,x<c\}$ whenever $I$ has points below $c$, ${\lim }_{x\to c^{+}}f(x)=\inf \{f(x):x\in I,x>c\}$ whenever it has points above $c$, and these satisfy ${\lim }_{x\to c^{-}}f(x)\le f(c)\le {\lim }_{x\to c^{+}}f(x)$ gives the one-sided limits at $c$, and the displayed equalities force $C_F(c^{-})=C_F(c)=C_F(c^{+})$. Hence $C_F$ is continuous at every interior point. Moreover, the displayed equalities show that $J_F$ has exactly the same left and right jumps as $F$. This proves claim 4 except for the tautological identity $F=J_F+C_F$.

To treat the endpoints, note first that step 2.1 with $x=a$ gives $0\le J_F(x)-{\beta }_a\le F(x)-{\lim }_{t\to a^{+}}F(t)$ for every $x>a$. Because the right-hand side tends to $0$ as $x\downarrow a$, $J_F(x)\to {\beta }_a$. Hence $C_F(x)=F(x)-J_F(x)⟶{\lim }_{x\to a^{+}}F(x)-{\beta }_a=F(a)=C_F(a),$ so $C_F$ is right-continuous at $a$. At $b$, the endpoint term in step 3.1 gives $J_F(b)-J_F(b^{-})=F(b)-F(b^{-}),$ and therefore $C_F(b^{-})=C_F(b)$. Thus $C_F$ is continuous on all of $[a,b]$. This proves claim 3.

Claim 4 contains the identity $F=J_F+C_F$ by definition of $C_F$. For claim 5, step 3.1 expresses $J_F(x)$ pointwise in terms of ${\beta }_a$, the interior left and right jump sizes of $F$, and the left jump at $b$, so those data determine $J_F$ uniquely. Once $J_F$ is fixed, the remainder is forced by $C_F=F-J_F$.

Steps 1.1 through 5.1 prove the theorem.