FALSE: every increasing function satisfies Newton-Leibniz with its derivative

Enumerate all closed rational intervals $I_n=[u_n,v_n]$ with $0\le u_n<v_n\le 1$. Let $c:[0,1]\to \mathbb{R}$ be the Cantor function of The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set. For each $n$, let $c_n$ be the function that is $0$ on $[0,u_n]$, is $1$ on $[v_n,1]$, and on $[u_n,v_n]$ is the affine rescaling of $c$. Then each $c_n$ is continuous, nondecreasing, and takes values in $[0,1]$. [given, choose] $S(x):={\sum }_{n\ge 1}{2}^{-n-1}{c}_n(x)\mathrm{.}$ The series converges uniformly because each summand is bounded by $2^{-n-1}$, so $S$ is continuous and nondecreasing.

If $x<y$, choose a rational interval $I_n$ with $x<u_n<v_n<y$. Then $c_n(x)=0$ and $c_n(y)=1$, so $S(y)-S(x)\ge 2^{-n-1}>0$. Hence $S$ is strictly increasing. For each $n$, the derivative of $c_n$ is $0$ almost everywhere because off the scaled Cantor set inside $I_n$ the function is locally constant by The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, and that scaled Cantor set is null because the Cantor set is null by The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points. Since each $c_n$ is nondecreasing, Fubini's theorem on term-by-term differentiation for pointwise sums of nondecreasing functions applies and gives $S^{\mathrm{\prime }}(x)={\sum }_{n\ge 1}{2}^{-n-1}{c}_n^{\mathrm{\prime }}(x)=0$ almost everywhere. Together with step 1.1, A singular function on a compact interval shows that $S$ is a singular function.

Therefore ${\int }_0^1{S}^{\mathrm{\prime }}(x)d\lambda (x)=0$ by Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree, while strict increase gives $S(1)-S(0)>0$. Hence Newton-Leibniz fails for this increasing function, and the statement is false.