If $f,g$ are integrable on $[a,b]$ then so are $\mid f\mid$, $f^2$, $fg$, $\max (f,g)$ and $\min (f,g)$, and $\mid {\int }_a^{b}f\mid \le {\int }_a^b\mid f\mid$

Let $a<b$ be reals and let $f,g:[a,b]\to \mathbb{R}$ be integrable (The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as ${\sup }_{P}L(f,P)$ and ${\inf }_{P}U(f,P)$, Darboux integrability as their equality, and the notation ${\int }_a^{b}f$). Then:

1. $\mathrm{\mid }f\mathrm{\mid }$, $f^2$ and $fg$ are integrable on $[a,b]$ (Absolute value in an ordered field, Integer powers $a^m$);
2. $\max (f,g)$ and $\min (f,g)$, defined pointwise (Maximum and minimum of a set), are integrable on $[a,b]$;
3. the triangle inequality for the integral: $\mid {\int }_a^{b}f\mid \le {\int }_a^b\mathrm{\mid }f\mathrm{\mid }\mathrm{.}$

Claim 3 is stated with $a<b$ and is not orientation-invariant. For $u>v$ the right-hand side is $-{\int }_v^u\mathrm{\mid }f\mathrm{\mid }\le 0$ while the left-hand side is $\ge 0$, so the inequality as written is false there. The form valid for every pair $u,v$ on which $f$ is integrable (The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$) is

$\mid {\int }_u^{v}f\mid \le \mid {\int }_u^v\mathrm{\mid }f\mathrm{\mid }\mid ,$

and that is the form the estimates below on this page use whenever the limits are not known to be in increasing order.

The converse of claim 1 fails. Integrability of $\mathrm{\mid }f\mathrm{\mid }$ does not give integrability of $f$; the witness is on the companion page.

• Every integrability clause comes from one theorem plus linearity. The only input that produces integrability is If $f$ is integrable on $[a,b]$ with values in $[m,M]$ and $\phi$ is continuous on $[m,M]$, then $\phi \circ f$ is integrable, with Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and ${\int }_a^b(\lambda f+\mu g)=\lambda {\int }_a^{b}f+\mu {\int }_a^{b}g$ recombining the pieces; claim 3 additionally uses If $f\le g$ on $[a,b]$ and both are integrable then ${\int }_a^{b}f\le {\int }_a^{b}g$; and $m(b-a)\le {\int }_a^{b}f\le M(b-a)$, which is the one place an inequality between integrals is needed. The identities of [L5] are algebra, and they are what turns a statement about composing with $\mathrm{\mid }\cdot \mathrm{\mid }$ and $(\cdot {)}^2$ into statements about products and lattice operations. In particular no new estimate on Darboux sums is made here.

• The polarisation identity is used, and it is why $f^2$ comes first. There is no direct route from integrability of $f$ and of $g$ to integrability of $fg$ through the composition theorem, because $(s,t)\mapsto st$ is a function of two variables and the theorem composes with one. Writing $fg$ through squares of sums and differences reduces it to the one-variable case.

• The inequality of claim 3 is the integral analogue of the triangle inequality, and like it, it can be strict: for $f(x)=x$ on $[-1,1]$ the left-hand side is $0$ and the right-hand side is $1$.

• Forward reference, orientation only. The witness refuting the converse of claim 1 is A function that is not Riemann integrable although $\mid f\mid$ is ↗ on the companion page; nothing above depends on it.