Basic Properties: Linearity and Additivity

Suppose you know that two functions are each Riemann integrable on an interval. Does it follow that their sum is integrable? Does the integral of the sum equal the sum of the integrals? This question sounds like it should have an immediate "yes". After all, integrals are built from sums of rectangle areas, and sums of sums are sums. But the Riemann integral is defined through a delicate limiting process. This process involves a supremum and an infimum over all partitions. Such quantities do not automatically play well with addition. The linearity of the integral is a theorem that must be proved. It is not an assumption we get for free. This matters because linearity and additivity are the load-bearing walls of integration theory. Nearly every computational technique rests on these two facts. This includes splitting an integral at a point, integrating a sum of functions term by term. It also includes pulling out constants and computing the integral of a piecewise-defined function. Without them, the Fundamental Theorem of Calculus would have no clean statement. The integral would be a curiosity rather than a computational tool. The big picture is that Riemann integrability is a robust property. It is preserved under the vector space operations of addition and scalar multiplication. It also interacts correctly with the geometric operation of subdividing the domain. Both facts trace back to the same technical machinery. This is the Cauchy Integrability Criterion, which characterizes integrability via upper and lower Darboux sums getting arbitrarily close together. It works together with the Equivalence of Riemann and Darboux Integrals. This equivalence lets us move freely between the Riemann-sum definition and the Darboux (upper/lower sum) definition. This article proves two central theorems rigorously. The first is linearity of the integral in the integrand (the integral of $\alpha f + \beta g$ equals $\alpha \int f + \beta \int g$ ). The second is additivity over adjacent intervals (splitting $\int_a^b$ into $\int_a^c + \int_c^b$ ). Both proofs use only the Cauchy criterion and elementary manipulations of suprema and infima. This keeps the logical chain transparent and self-contained.

Key takeaways

If $f,g$ are Riemann integrable on $[a,b]$ and $\alpha,\beta \in \mathbb{R}$ , then $\alpha f + \beta g$ is integrable and $\int_a^b (\alpha f+\beta g) = \alpha\int_a^b f + \beta \int_a^b g$ .
If $f$ is integrable on $[a,b]$ and $c \in [a,b]$ , then $f$ is integrable on $[a,c]$ and $[c,b]$ , and $\int_a^b f = \int_a^c f + \int_c^b f$ .
The proofs use the Cauchy Integrability Criterion: integrability $\iff$ upper and lower sums can be made arbitrarily close.
A key subtlety: $\sup(f+g) \le \sup f + \sup g$ in general, with strict inequality possible — this is why linearity needs a real proof.
Additivity requires refining partitions to include the splitting point $c$ , since upper/lower sums are only well-behaved for partitions that "see" $c$ .
These two results are the algebraic backbone used immediately afterward to prove monotonicity, the triangle inequality for integrals, and the Fundamental Theorem of Calculus.
Key idea: integrability is stable under linear combination and under domain-splitting, and both stabilities are proved by controlling upper/lower Darboux sums via the Cauchy criterion.

Where this lives today

Linearity of the integral is the algebraic engine behind term-by-term integration of power series and Fourier series, where an infinite sum is integrated by justifying (via uniform convergence) that the linearity property extends from finite sums to the limit. In probability and statistics, linearity of expectation — $E[\alpha X + \beta Y] = \alpha E[X] + \beta E[Y]$ — is the exact measure-theoretic analogue of this theorem, proved by essentially the same squeezing technique on integrals against a probability measure. In numerical analysis, additivity over adjacent intervals underlies composite quadrature rules (composite Simpson's rule, adaptive quadrature), where an integral over a large interval is approximated by summing accurate approximations on many small subintervals. In physics and engineering, additivity justifies computing work, charge, or probability by integrating piecewise over regions with different physical laws (e.g., different force laws in different zones) and adding the contributions — exactly the logic of Example 2 above. In signal processing, linearity of the integral (as a linear functional) is precisely what makes convolution and Fourier transforms linear operations, connecting back to Equivalence of Riemann and Darboux Integrals as the rigorous foundation making such operations well-defined.

:::quiz[If $f$ and $g$ are each Riemann integrable on $[0,1]$ , which statement is guaranteed true by the Linearity Theorem?]

$\int_0^1 fg = \left(\int_0^1 f\right)\left(\int_0^1 g\right)$
$\int_0^1 (3f - 2g) = 3\int_0^1 f - 2\int_0^1 g$ ✓
$\sup(f+g) = \sup f + \sup g$ on every subinterval
$f/g$ is automatically Riemann integrable

:::quiz[If $f$ is integrable on $[0,5]$ , which of the following is proved by the Additivity Theorem?]

$f$ is integrable on $[0,2]$ and $[2,5]$ , with $\int_0^5 f = \int_0^2 f + \int_2^5 f$ ✓
$f$ must be continuous at $x=2$
$\int_0^5 f = \int_0^2 f \cdot \int_2^5 f$
$f$ restricted to $[0,2]$ has the same supremum as $f$ on $[0,5]$

:::faq[Can I split an integral at a point outside $[a,b]$ , like using $c > b$ ?] Yes, provided you use the signed-interval convention $\int_p^q f = -\int_q^p f$ for $p > q$ , and $f$ is integrable on the larger interval containing $a,b,c$ . For instance, if $f$ is integrable on $[a,c]$ with $b\in(a,c)$ , the Additivity Theorem directly gives $\int_a^c f = \int_a^b f + \int_b^c f$ , which rearranges to $\int_a^b f = \int_a^c f - \int_b^c f = \int_a^c f + \int_c^b f$ , matching the general corollary. The convention is what makes the additivity formula work uniformly regardless of the relative order of $a,b,c$ , and it is essential for later results like the Fundamental Theorem of Calculus, where the upper limit of integration is allowed to vary freely.

Ask the tutor: Why does the proof of linearity break down if we only know $f+g$ is integrable, without knowing $f$ and $g$ are integrable separately?

f:[a,b]\to\mathbb{R}

Intuition

Think of $\int_a^b f$ as a signed area, built by squeezing it between lower staircase approximations (using infima on each subinterval) and upper staircase approximations (using suprema). Integrability means these staircases can be forced together as tightly as we like. Now imagine two functions $f$ and $g$ , each squeezed by their own staircases. When we add them, the pointwise sum $f+g$ certainly gets squeezed between $L(f,P)+L(g,P)$ and $U(f,P)+U(g,P)$ on any common partition $P$ — but the subtlety is that the supremum of $f+g$ on a subinterval need not equal $\sup f + \sup g$ on that subinterval; it can be strictly smaller, because $f$ and $g$ might achieve their maxima at different points. Linearity survives this gap only because integrability lets us squeeze $U$ and $L$ together independently for $f$ and for $g$ , and then use inequalities (not equalities) in the right direction to sandwich $\int(f+g)$ . For additivity, picture the interval $[a,b]$ with a marked point $c$ . A partition of $[a,b]$ that happens to include $c$ as a partition point splits cleanly into a partition of $[a,c]$ and one of $[c,b]$ , and the upper/lower sums add. The only work is showing that refining any partition to include $c$ never hurts our control on the sums — refinement never increases upper sums and never decreases lower sums, a fact borrowed from the Cauchy criterion machinery.

Figure

Splitting a partition at the point

c

f,g:[a,b]\to\mathbb{R}

f,g:[a,b]\to\mathbb{R}

[x_{i-1},x_i]

f+g

f:[a,b]\to\mathbb{R}

\Rightarrow

f

Worked examples

Example 1 (Linearity with asymmetric functions and irrational coefficients). Let $f(x) = x^2$ and $g(x) = \lfloor x \rfloor$ (floor function) on $[0,2]$ , and let $\alpha = \sqrt{2}$ , $\beta = -3$ . Checking hypotheses. $f(x)=x^2$ is continuous on $[0,2]$ , hence Riemann integrable (continuous functions on closed bounded intervals are integrable — a standard consequence of uniform continuity and the Cauchy criterion). $g(x) = \lfloor x\rfloor$ is a step function with a single jump at $x=1$ ; it is bounded ( $0\le g\le 2$ ) and has only one discontinuity, so it is Riemann integrable (monotone/step functions with finitely many discontinuities are integrable). Both hypotheses of the Linearity Theorem are satisfied. Computing each integral. By the Fundamental Theorem (used here only for arithmetic, not to justify the structural theorem itself):

\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}.

For $g$ : on $[0,1)$ , $g\equiv 0$ ; on $[1,2)$ , $g\equiv 1$ ; at the single point $x=2$ , $g(2)=2$ , which does not affect the integral (a single point has no effect on Darboux sums). So

\int_0^2 \lfloor x\rfloor\,dx = \int_0^1 0\,dx + \int_1^2 1\,dx = 0 + 1 = 1.

Applying linearity. The theorem guarantees $\sqrt2\, f - 3g$ is integrable on $[0,2]$ and

\int_0^2 \left(\sqrt2\,x^2 - 3\lfloor x\rfloor\right)dx = \sqrt2 \int_0^2 x^2\,dx - 3\int_0^2 \lfloor x\rfloor\,dx = \sqrt2 \cdot \frac{8}{3} - 3\cdot 1 = \frac{8\sqrt2}{3} - 3.

Numerically, $\frac{8\sqrt2}{3}\approx 3.7712$ , so the value is approximately $0.7712$ . Note that computing $\int (\sqrt2 f - 3g)$ directly from Darboux sums of the combined function would be far messier, since $\sqrt2 x^2 - 3\lfloor x \rfloor$ has an awkward jump; linearity lets us bypass that entirely by working with $f$ and $g$ separately.

Example 2 (Additivity to compute an integral of a genuinely piecewise function). Define $h:[-1,3]\to\mathbb{R}$ by

h(x) = \begin{cases} x^2 + 1, & -1\le x < 1,\\ 4 - x, & 1 \le x \le 3.\end{cases}

Checking hypotheses. On $[-1,1]$ , redefine $\tilde h_1(x)=x^2+1$ everywhere (agrees with $h$ on $[-1,1)$ and we assign $\tilde h_1(1)=2$ ); this is continuous, hence integrable on $[-1,1]$ , and changing the value at the single point $x=1$ does not change the integral. On $[1,3]$ , $h(x)=4-x$ is continuous, hence integrable. So by the Additivity Theorem (applied at the splitting point $c=1\in(-1,3)$ ), $h$ is integrable on $[-1,3]$ with

\int_{-1}^{3} h = \int_{-1}^{1} (x^2+1)\,dx + \int_{1}^{3} (4-x)\,dx.

Computing each piece.

\int_{-1}^1 (x^2+1)\,dx = \left[\frac{x^3}{3}+x\right]_{-1}^1 = \left(\frac13+1\right)-\left(-\frac13-1\right) = \frac{4}{3}+\frac{4}{3}\cdot\frac12\cdot 2

Let us instead compute carefully: at $x=1$ : $\frac13+1=\frac43$ . At $x=-1$ : $-\frac13-1=-\frac43$ . Difference: $\frac43-(-\frac43)=\frac83$ .

\int_1^3 (4-x)\,dx = \left[4x - \frac{x^2}{2}\right]_1^3 = \left(12-\frac92\right)-\left(4-\frac12\right) = \frac{15}{2} - \frac{7}{2} = 4.

Total.

\int_{-1}^3 h = \frac83 + 4 = \frac{20}{3} \approx 6.667.

This illustrates the additivity theorem exactly as intended: a function defined by different rules on two adjacent pieces is integrated piece by piece, and the theorem is what licenses gluing the results together into a single well-defined integral over $[-1,3]$ .

U(f,P)

Scope, edge cases & common traps

Situation	What holds	What can fail
$f,g$ bounded and integrable on $[a,b]$	Linearity holds exactly as stated	—
$f$ or $g$ unbounded	Integral undefined for that function	Linearity statement is meaningless, not merely false
$c \notin [a,b]$	Additivity formula still makes sense using signed-interval convention	Must not naively assume $c\in(a,b)$ ; check the corollary's convention
$f$ integrable on $[a,c]$ , $[c,b]$ separately but defined differently at $x=c$	Still integrable on $[a,b]$ ; value at one point never affects the integral	Beginners sometimes worry the "jump" at $c$ breaks integrability — it does not
Infinitely many splitting points (e.g. splitting into infinitely many subintervals)	Additivity as stated is only proved for finitely many splits	Passing to infinite sums requires separate justification (uniform convergence / dominated convergence type arguments), not covered by this theorem

Common trapThe classic error: assuming

\sup(f+g) = \sup f + \sup g

A very common flawed "proof" of linearity asserts that upper sums add exactly, i.e. $U(f+g,P) = U(f,P)+U(g,P)$ , by claiming the supremum of a sum is the sum of suprema. This is false in general: take $f(x) = \sin(2\pi x)$ and $g(x) = \cos(2\pi x)$ on $[0,1]$ ; each has supremum $1$ on the whole interval, but $f+g = \sqrt2 \sin(2\pi x + \pi/4)$ has supremum $\sqrt2 < 1+1=2$ . The correct statement is only the inequality $\sup(f+g) \le \sup f + \sup g$ (and symmetrically $\inf(f+g)\ge \inf f + \inf g$ ), and the entire proof of linearity is built around squeezing with this one-directional inequality plus the Cauchy criterion — not around a false equality of suprema.

The mathematician behind it

Origins · 1 minBernhard Riemann and how this was discovered

Bernhard Riemann

1826–1866 · Göttingen, mid-19th century analysis

Riemann introduced the rigorous definition of the definite integral bearing his name in his 1854 habilitation lecture on trigonometric series, seeking to understand exactly which functions could be represented by Fourier series. His formulation — via limits of sums over partitions with arbitrary tags — was later reorganized by Gaston Darboux using upper and lower sums, giving the equivalent and computationally convenient formulation used throughout this article. Riemann's broader legacy spans complex analysis, differential geometry (Riemannian manifolds), and the Riemann Hypothesis in number theory, but his integral remains the entry point through which nearly every student first encounters rigorous integration theory.

History

- 1854: Riemann defines the integral via limits of sums in his habilitation thesis on trigonometric series, first giving a precise integrability criterion.

- 1875: Gaston Darboux reformulates Riemann's integral using upper and lower sums, yielding the modern Darboux/Riemann equivalence used pedagogically today.

- 1881: Vito Volterra constructs pathological examples clarifying the boundary between Riemann integrability and mere boundedness, sharpening understanding of when linearity-type arguments apply.

- 1902: Henri Lebesgue introduces measure-theoretic integration, subsuming Riemann's theory and explaining exactly which bounded functions are Riemann integrable (those continuous almost everywhere).

- 1920s–on: Linearity and additivity of the integral become foundational lemmas in every rigorous calculus and real analysis curriculum, stated and proved essentially as here before the Fundamental Theorem of Calculus.

Why doesn't

U(f+g,P) = U(f,P) + U(g,P)

hold in general?

Because the supremum of a sum of two functions on a set need not equal the sum of their individual suprema. On a given subinterval, $f$ might achieve its maximum at one point while $g$ achieves its maximum at a different point, so $f+g$ can never actually reach the sum of the two individual maxima simultaneously. The only guaranteed relationship is the inequality $\sup(f+g)\le \sup f+\sup g$ , obtained by noting $f(x)+g(x)\le \sup f+\sup g$ for every $x$ , so the right side is an upper bound (though not necessarily the least one) for $f+g$ . This is precisely why the proof of linearity must proceed via inequalities and the Cauchy criterion, squeezing $\int(f+g)$ between $L(f,P)+L(g,P)$ and $U(f,P)+U(g,P)$ , rather than via a direct equality of upper sums.

Does linearity extend to infinite sums, like

\int \sum_{n=1}^\infty f_n

Not automatically. The theorem proved here is strictly about finite linear combinations: $\alpha f+\beta g$ , or by induction any finite sum $\sum_{i=1}^n \alpha_i f_i$ . Extending to an infinite series requires additional hypotheses, typically that the partial sums converge uniformly to the limit function on $[a,b]$ ; uniform convergence is exactly the condition that lets you interchange the limit (infinite sum) and the integral, using an $\varepsilon/3$ -style argument built on top of finite linearity. Without uniform convergence, term-by-term integration can fail — there are pointwise-convergent sequences of integrable functions whose integrals do not converge to the integral of the limit.

Is Riemann integrability itself needed, or would just boundedness suffice for these theorems?

Boundedness alone is not enough; genuine Riemann integrability of each piece is required. There exist bounded functions that are not Riemann integrable — the classic example is the Dirichlet function (indicator of the rationals) on $[0,1]$ , for which every upper sum equals $1$ and every lower sum equals $0$ , so the Cauchy criterion fails and no integral exists. If $f$ is such a non-integrable bounded function and $g = -f$ , then $f+g \equiv 0$ is trivially integrable, but neither $f$ nor $g$ individually has an integral to plug into a linearity formula. The theorems in this article are conditional statements: given integrability of the pieces, conclude integrability and the formula for the combination — they say nothing about combinations of non-integrable pieces.

Why is the Cauchy Integrability Criterion the right tool for these proofs, rather than working directly with Riemann sums?

The Cauchy criterion reduces integrability to a single, checkable inequality — the existence of a partition making $U(f,P)-L(f,P)$ arbitrarily small — which interacts very cleanly with algebraic operations because $U$ and $L$ are built from suprema and infima on subintervals, for which we have clean (if one-directional) inequalities under addition and scaling. Working directly with tagged Riemann sums and mesh-limits is possible but more cumbersome here, since one would need to carefully synchronize tag choices and mesh sizes between $f$ and $g$ . The Equivalence of Riemann and Darboux Integrals guarantees both approaches give the same integral, so we are free to use whichever is more convenient — and for algebraic properties like linearity and additivity, the Darboux/Cauchy approach is decisively more convenient.

Does additivity mean I can split at any finite number of points, not just one?

Yes — this follows by induction from the two-point case proved here. If $f$ is integrable on $[a,b]$ and $a = c_0 < c_1 < \cdots < c_n = b$ , then applying the Additivity Theorem repeatedly (first splitting at $c_1$ , then splitting the piece $[c_1,b]$ at $c_2$ , and so on) shows $f$ is integrable on each $[c_{i-1},c_i]$ and

\int_a^b f = \sum_{i=1}^n \int_{c_{i-1}}^{c_i} f.

This finite generalization is exactly what licenses computing the integral of a piecewise-defined function with finitely many pieces, as in Example 2, and it is the version of additivity most often invoked in practice.

How do linearity and additivity foreshadow the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus typically defines $F(x) = \int_a^x f(t)\,dt$ and studies its derivative. Additivity is exactly what makes $F$ well-defined and lets you compute $F(x+h)-F(x) = \int_x^{x+h} f(t)\,dt$ as a single small integral rather than a difference of two large ones — this identity is a direct instance of the additivity theorem. Linearity, meanwhile, is what lets you differentiate $F$ termwise when $f$ itself is built from sums of simpler functions, and combined with the forthcoming Monotonicity and Triangle Inequality for Integrals, it supplies the estimates needed to squeeze the difference quotient of $F$ and prove $F'(x)=f(x)$ at points of continuity.