If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$; and $m(b-a) \le \int_a^b f \le M(b-a)$

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. Nonnegativity. If $f(x) \ge 0$ for every $x \in [a,b]$ then $\int_a^b f \ge 0$.
2. Monotonicity. If $f(x) \le g(x)$ for every $x \in [a,b]$ then $\int_a^b f \;\le\; \int_a^b g .$
3. Two-sided bound. If $m \le f(x) \le M$ for every $x \in [a,b]$, with $m, M$ real, then $m\,(b-a) \;\le\; \int_a^b f \;\le\; M\,(b-a) .$

Equality in claim 1 does not force $f$ to vanish. A nonnegative integrable function with integral $0$ may be positive at infinitely many points; that is FALSE: a nonnegative Riemann integrable function on $[a,b]$ with $\int_a^b f = 0$ is identically zero on the previous page's companion. Under the additional hypothesis of continuity the conclusion does hold, and that is A continuous $f \ge 0$ on $[a,b]$ with $\int_a^b f = 0$ is identically $0$ below.

Claim 2 is stated for $a < b$ and is not orientation-invariant. With the convention of The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$, $f \le g$ gives $\int_u^v f \le \int_u^v g$ when $u \le v$ and the reverse inequality when $u \ge v$, since both sides change sign together.

• Claim 3 is cited, not reproved. If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$ already proves the five-term chain for every partition, and it is the item that also computes the integral of a constant, $\int_a^b c = c(b-a)$. Claim 3 is that chain read at an integrable $f$; nothing new is established here.

• Claim 2 is proved through claim 1 and linearity, and not by comparing Darboux sums. Comparing sums works too, since $f \le g$ gives $m_i(f) \le m_i(g)$ and $M_i(f) \le M_i(g)$ on every subinterval, but the route through $g - f$ is shorter and uses only results already available. Either way the hypothesis is a pointwise inequality on the whole of $[a,b]$; an inequality holding off a finite set gives the same conclusion, by Changing an integrable function at finitely many points changes neither its integrability nor its integral, and that is a separate statement.

• What claim 2 is for. It is what turns a pointwise estimate on an integrand into an estimate on the integral, and every estimate of that shape on this page and its companion is an application of it. No count of those applications is asserted here; the dependency graph of the page is where that is read off.