Proof of Product Rule and Reflection for Indefinite Riemann Integrals

Claim 1. The functions $F$ and $G$ are continuous on $[a,b]$ by claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals (a constant plus a continuous indefinite integral), and then $h:=Fg+fG$ is continuous by Continuity of Sums and Products of Real-Valued Functions on a Metric Space.

By claim 3 of Fundamental Theorem of Calculus, Part I, on a Closed Real Interval, the function $t\mapsto\int_a^t f(r)\,dr$ is differentiable with derivative $f$ at every point of $(a,b)$, that is, at every interior point of $[a,b]$. Adding the constant $c_F$ does not change difference quotients, so $F$ is differentiable with derivative $f$ at every interior point of $[a,b]$, and likewise $G$ with derivative $g$. By the product rule of Sum and Product Rules for One-Dimensional Derivatives and Continuity, applied at each point of $(a,b)$, $H:=FG$ is differentiable there with derivative $h$; and $H$ is continuous on $[a,b]$ by Continuity of Sums and Products of Real-Valued Functions on a Metric Space.

Fix $t\in[a,b]$. If $t=a$, both sides equal $c_Fc_G$ by the degenerate-interval convention. Assume $t>a$. For natural $n\ge3$ put $a'_n:=a+(t-a)/n$ and $t'_n:=t-(t-a)/n$, so that $a<a'_n<t'_n<t\le b$. By Restriction Stability of Continuity and of the Derivative, the restriction of $H$ to $[a'_n,t'_n]$ is continuous there and differentiable at every point of $(a'_n,t'_n)\subseteq(a,b)$ with derivative $h$, and the restriction of $h$ to $[a'_n,t'_n]$ is likewise continuous there by claim 1 of that lemma, hence Riemann integrable on $[a'_n,t'_n]$ by claim 3 of the integral toolkit on a compact interval. Hence Fundamental Theorem of Calculus, Part II, on a Closed Real Interval gives

$$H(t'_n)-H(a'_n)=\int_{a'_n}^{t'_n}h(r)\,dr=\int_a^{t'_n}h(r)\,dr-\int_a^{a'_n}h(r)\,dr ,$$

the second equality by claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals. As $n\to\infty$, $a'_n\to a$ and $t'_n\to t$; the left-hand side converges to $H(t)-H(a)$ by the continuity of $H$, and the right-hand side converges to $\int_a^t h(r)\,dr-0$ by the continuity of $s\mapsto\int_a^s h(r)\,dr$ on $[a,b]$ (claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals). Hence $H(t)-H(a)=\int_a^t h(r)\,dr$, and since $H(a)=F(a)G(a)=c_Fc_G$ (the integrals over $[a,a]$ vanish), this is exactly the asserted identity.

Claim 2. The map $\rho\mapsto a+b-\rho$ is continuous from $[a,b]$ to $[a,b]$ (built from constants and the identity by Continuity of Sums and Products of Real-Valued Functions on a Metric Space), and $\psi:=\varphi(a+b-\cdot)$ is continuous as a composition of continuous maps, by Composition of Continuous Euclidean Maps, passing between that statement and continuity relative to the interval by Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions.

Fix $t\in[a,b]$; if $t=b$ both sides vanish by the degenerate convention, so assume $t<b$. Both $\varphi$ on $[t,b]$ and $\psi$ on $[a,a+b-t]$ are continuous, hence Riemann integrable, by claim 3 of the integral toolkit on a compact interval. Write $I:=\int_t^b\varphi(r)\,dr$.

Let $\varepsilon>0$ and let $\delta>0$ be as in the defining condition of Riemann Integrability on a Closed Interval for $\varphi$ on $[t,b]$, $I$, and $\varepsilon$: every tagged partition of $[t,b]$ with mesh less than $\delta$ has Riemann sum within $\varepsilon$ of $I$. Let now $(P,\tau)$ be any tagged partition of $[a,a+b-t]$ with mesh less than $\delta$, say with partition points $a=y_0<y_1<\dots<y_n=a+b-t$ and tags $\tau_p\in[y_{p-1},y_p]$. Define its reflection: the points $z_p:=a+b-y_{n-p}$ ($0\le p\le n$) satisfy $t=z_0<z_1<\dots<z_n=b$, and the tags $\sigma_p:=a+b-\tau_{n-p+1}$ satisfy $\sigma_p\in[z_{p-1},z_p]$, since $\rho\mapsto a+b-\rho$ is decreasing and maps $[y_{n-p},y_{n-p+1}]$ onto $[z_{p-1},z_p]$. The subinterval lengths agree, $z_p-z_{p-1}=y_{n-p+1}-y_{n-p}$, so the reflection is a tagged partition of $[t,b]$ with the same mesh, less than $\delta$. Its Riemann sum equals that of $(P,\tau)$: reindexing by $q=n-p+1$,

$$\sum_{p=1}^{n}\varphi(\sigma_p)\,(z_p-z_{p-1})=\sum_{q=1}^{n}\varphi(a+b-\tau_{q})\,(y_{q}-y_{q-1})=\sum_{q=1}^{n}\psi(\tau_{q})\,(y_{q}-y_{q-1}).$$

Hence every tagged partition of $[a,a+b-t]$ with mesh less than $\delta$ has Riemann sum within $\varepsilon$ of $I$. Thus the number $I$ satisfies the defining condition of the Riemann integral for $\psi$ on $[a,a+b-t]$; and this number is unique, since two numbers each within $\varepsilon$ of the Riemann sums of arbitrarily fine tagged partitions differ by less than $2\varepsilon$ for every $\varepsilon>0$. Therefore $\int_a^{a+b-t}\psi(\rho)\,d\rho=I$, which is the asserted identity. $\square$