Zero-start integral inequality

For a continuous function on $[a,b]$, the Darboux definition gives $m(b-a)\le {\int }_a^{b}f\le M(b-a)$ whenever $m\le f\le M$, because every lower and upper sum has these bounds. For $a<c<b$, concatenating partitions of $[a,c]$ and $[c,b]$ adds both their lower sums and their upper sums. By integrability [F1], choose on each interval a lower sum within $\eta$ of its integral and an upper sum within $\eta$ of its integral, and refine these finitely many partitions together. The concatenated sums sandwich ${\int }_a^{b}f$ within $2\eta$ of ${\int }_a^{c}f+{\int }_c^{b}f$; letting $\eta \downarrow 0$ proves additivity. This also holds when $c$ is an endpoint under the zero-interval convention. Define reversed integrals by ${\int }_b^{a}f=-{\int }_a^{b}f$. By [F1], the integral defining $h(t)={\int }_0^{t}f(s)ds$ exists. Continuity of $f$ gives $h^{\mathrm{\prime }}(t)=f(t)$ on $(0,T)$: the difference quotient error is bounded by ${\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le \mathrm{\mid }r\mathrm{\mid }}\mathrm{\mid }f(s)-f(t)\mathrm{\mid }$, which tends to zero as $r\to 0$. By [F4], choose a finite $M$ with $0\le f\le M$. Then $\mathrm{\mid }h(t)-h(s)\mathrm{\mid }=\mathrm{\mid }{\int }_s^{t}f\mathrm{\mid }\le M\mathrm{\mid }t-s\mathrm{\mid }$ for all $s,t\in [0,T]$, which proves continuity at the endpoints as well as in the interior. The hypothesis gives $h^{\mathrm{\prime }}\le Lh$, so [F2]–[F3] yield $g^{\mathrm{\prime }}(t)\le 0$ for $g(t)=e^{-Lt}h(t)$ on $(0,T)$. The mean value theorem makes $g$ nonincreasing; since $g(0)=0$ and $g\ge 0$, $g\equiv 0$ and $h\equiv 0$.

For any $t>0$, the nonnegative continuous function $f$ has integral $h(t)=0$. If $f(t_0)>0$ at any $t_0>0$, continuity gives a positive lower bound on a nonempty subinterval of $[0,t_0]$, contradicting $h(t_0)=0$. At $t=0$ the assumed inequality gives $f(0)\le 0$; nonnegativity gives $f(0)=0$. Hence $f\equiv 0$.