For a nondecreasing function, the derivative is measurable and integrable and its integral is bounded by the total increase

By A monotone function is differentiable almost everywhere by the rising-sun route, $F^{\mathrm{\prime }}$ exists almost everywhere on $(a,b)$. Extend $F$ to $[a,b+1]$ by setting $\tilde{F}(x)=F(x)$ for $x\le b$ and $\tilde{F}(x)=F(b)$ for $x\ge b$. For each $n\ge 1$ define $h_n(x):=n(\tilde{F}(x+1\mathrm{/}n)-\tilde{F}(x))$ for $x\in [a,b]$. Because monotone functions are Borel measurable (Every monotone real function is Borel measurable) and arithmetic preserves measurability (Arithmetic and lattice operations preserve measurability whenever they are defined), each $h_n$ is measurable and nonnegative. At every point where $F^{\mathrm{\prime }}$ exists, $h_n(x)\to F^{\mathrm{\prime }}(x)$.

For every $n$, ${\int }_a^b{h}_n(x)d\lambda (x)=n{\int }_a^b\tilde{F}(x+1\mathrm{/}n)d\lambda (x)-n{\int }_a^b\tilde{F}(x)d\lambda (x)$. After the change of variable $t=x+1\mathrm{/}n$ in the first integral, this becomes ${\int }_a^b{h}_n=n{\int }_b^{b+1\mathrm{/}n}\tilde{F}(t)dt-n{\int }_a^{a+1\mathrm{/}n}F(t)dt\le F(b)-F(a)$, because $\tilde{F}(t)\le F(b)$ on $[b,b+1\mathrm{/}n]$ and $F(t)\ge F(a)$ on $[a,a+1\mathrm{/}n]$.

Fatou's lemma Fatou's lemma gives ${\int }_a^b{F}^{\mathrm{\prime }}(x)d\lambda (x)\le {\int }_a^b{\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{h}_n(x)d\lambda (x)\le {\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{\int }_a^b{h}_n(x)d\lambda (x)\le F(b)-F(a)$. Thus $F^{\mathrm{\prime }}$ is integrable and obeys the claimed bound. Since it is almost everywhere the pointwise limit of the measurable functions $h_n$, it is measurable as well, after changing it on the null exceptional set if needed and using Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree.

Steps 1.1 through 3.1 prove the theorem.