Deriving Linearity From Riemann Sums

From Riemann Sums to Linearity

A definite integral is defined through limits of finite sums. This makes certain rules for integrals more than convenient algebra: they follow from exact identities that hold before any limit is taken.

For example, multiplying every function value in a Riemann sum by a number $c$ multiplies the whole sum by $c$. Likewise, the Riemann sum for $f+g$ is the sum of the Riemann sums for $f$ and $g$, as long as the same partition and tags are used. To turn these observations into rules for integrals, we must also know that the relevant functions are integrable. The limiting argument supplies both the rules and their hypotheses.

Tagged Sums and Integrability

Let $a<b$. A partition of $[a,b]$ is a finite sequence $a=x_0<x_1<\cdots<x_n=b$. In each subinterval $[x_{i-1},x_i]$, choose a tag $\xi_i$. The resulting tagged Riemann sum for a function $h$ is

S(h;P,\xi) = \sum_{i=1}^{n}h(\xi_i)\Delta x_i, \qquad \Delta x_i=x_i-x_{i-1}.

The mesh of the partition is the width of its largest subinterval:

\|P\|=\max_{1\leq i\leq n}\Delta x_i.

In this tutorial, Riemann integrability means that $h$ is bounded on $[a,b]$ and there is a number $I$ such that, for every $\varepsilon>0$, some $\delta>0$ has the following property: every tagged partition with $\|P\|<\delta$ satisfies $|S(h;P,\xi)-I|<\varepsilon$. The number $I$ is the definite integral $\int_a^b h(x)\,dx$.

Definition: A bounded function $h$ is Riemann integrable on $[a,b]$ if all its tagged Riemann sums approach one common limit as the mesh tends to zero. The condition applies to every choice of tags, not just to one convenient sequence of sums.

That last point matters. Calculating a limit for one particular choice of partitions and tags does not, by itself, establish Riemann integrability. The definition requires the same limit for all sufficiently fine tagged partitions.

The Exact Identities Behind the Rules

Fix any partition and any choice of tags. Since the subinterval widths do not depend on the function, ordinary distributivity gives the identities

S(cf;P,\xi)=cS(f;P,\xi)

and

S(f+g;P,\xi)=S(f;P,\xi)+S(g;P,\xi).

For the second identity, the same tags must be used in all three sums. Indeed, the $i$-th contribution on the left is $(f(\xi_i)+g(\xi_i))\Delta x_i$, which equals $f(\xi_i)\Delta x_i+g(\xi_i)\Delta x_i$. Summing over all subintervals proves the identity exactly.

Theorem: If $f$ is Riemann integrable on $[a,b]$, then $cf$ is Riemann integrable for every real number $c$, and $$ \int_a^b cf(x)\,dx=c\int_a^b f(x)\,dx. $$ If $f$ and $g$ are both Riemann integrable on $[a,b]$, then $f+g$ is Riemann integrable and $$ \int_a^b (f(x)+g(x))\,dx = \int_a^b f(x)\,dx+\int_a^b g(x)\,dx. $$

Why the Rules Hold

The proof uses the exact sum identities and the definition of a limit of tagged sums. For scalar multiplication, first suppose $c\neq 0$, and write $I=\int_a^b f(x)\,dx$. Given $\varepsilon>0$, integrability of $f$ gives a $\delta>0$ such that every tagged partition with mesh below $\delta$ satisfies

|S(f;P,\xi)-I|<\frac{\varepsilon}{|c|}.

For any such partition, the exact identity for tagged sums now gives

|S(cf;P,\xi)-cI| = |cS(f;P,\xi)-cI| = |c|\,|S(f;P,\xi)-I| < \varepsilon.

Thus every sufficiently fine tagged sum for $cf$ approaches $cI$, so $cf$ is integrable and its integral is $cI$. If $c=0$, every tagged sum for $cf$ is zero, so the same conclusion holds directly.

For the sum rule, let $I_f=\int_a^b f(x)\,dx$ and $I_g=\int_a^b g(x)\,dx$. Given $\varepsilon>0$, choose $\delta_f>0$ and $\delta_g>0$ from the definitions of integrability so that sums for $f$ differ from $I_f$ by less than $\varepsilon/2$ when the mesh is below $\delta_f$, and sums for $g$ differ from $I_g$ by less than $\varepsilon/2$ when the mesh is below $\delta_g$. Set $\delta=\min(\delta_f,\delta_g)$. For any tagged partition with $\|P\|<\delta$, use the same tags for $f$, $g$, and $f+g$. Then

\begin{aligned} |S(f+g;P,\xi)-(I_f+I_g)| &=|S(f;P,\xi)-I_f+S(g;P,\xi)-I_g|\\ &\leq |S(f;P,\xi)-I_f|+|S(g;P,\xi)-I_g|\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. \end{aligned}

Therefore $f+g$ is integrable, with integral $I_f+I_g$. This proves the rule, including the integrability claim, rather than assuming that the sum is integrable in advance.

Consequence: For Riemann-integrable $f$ and $g$, and real numbers $\alpha$ and $\beta$, the function $\alpha f+\beta g$ is Riemann integrable and $$ \int_a^b (\alpha f(x)+\beta g(x))\,dx = \alpha\int_a^b f(x)\,dx+\beta\int_a^b g(x)\,dx. $$ This follows by applying the scalar-multiple rule and then the sum rule.

Worked Examples

Worked Example: Scaling a Riemann Sum

Consider $f(x)=x$ on $[0,2]$, and use the uniform partition into $n$ subintervals with right-endpoint tags. Each subinterval has width $\Delta x=2/n$, and its $i$-th right endpoint is $x_i=2i/n$. The tagged sum for $f$ is

\begin{aligned} S(f;P,\xi) &=\sum_{i=1}^{n}\frac{2i}{n}\frac{2}{n}\\ &=\frac{4}{n^2}\sum_{i=1}^{n}i\\ &=\frac{4}{n^2}\frac{n(n+1)}{2}\\ &=\frac{2(n+1)}{n} =2+\frac{2}{n}. \end{aligned}

As $n\to\infty$, the mesh $2/n$ tends to zero and the sums tend to $2$, so $\int_0^2 x\,dx=2$. For the function $3f$, the tagged-sum identity gives

S(3f;P,\xi)=3S(f;P,\xi) =6+\frac{6}{n} \longrightarrow 6.

Thus $\int_0^2 3x\,dx=6$, which agrees with $3\int_0^2 x\,dx=3\cdot2=6$. As a check, the finite sums also satisfy $6+6/n=3(2+2/n)$ for every positive integer $n$, not only in the limit.

Worked Example: Adding Two Integrands

On $[0,1]$, take $f(x)=x$ and $g(x)=x^2$, again using a uniform partition and right-endpoint tags. Here $\Delta x=1/n$ and $x_i=i/n$. First compute each sum:

\begin{aligned} S(f;P,\xi) &=\sum_{i=1}^{n}\frac{i}{n}\frac{1}{n} =\frac{1}{n^2}\frac{n(n+1)}{2} =\frac{n+1}{2n},\\ S(g;P,\xi) &=\sum_{i=1}^{n}\frac{i^2}{n^2}\frac{1}{n} =\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6} =\frac{(n+1)(2n+1)}{6n^2}. \end{aligned}

The sum identity applies to these same tags. Therefore

\begin{aligned} S(f+g;P,\xi) &=\frac{n+1}{2n}+\frac{(n+1)(2n+1)}{6n^2}\\ &=\frac{3n(n+1)+(n+1)(2n+1)}{6n^2}\\ &=\frac{(n+1)(5n+1)}{6n^2}. \end{aligned}

Taking limits, the first sum tends to $1/2$, the second to $1/3$, and their sum to $5/6$. Hence

\int_0^1(x+x^2)\,dx = \frac{5}{6} = \frac{1}{2}+\frac{1}{3} = \int_0^1x\,dx+\int_0^1x^2\,dx.

For a finite-sum check, the expression for $S(f+g;P,\xi)$ is exactly the sum of the two displayed component sums; both simplify to $(n+1)(5n+1)/(6n^2)$. The integral identity follows because each component is integrable and the corresponding sums converge.

Worked Example: Cancellation Can Restore Integrability

The assumptions in the sum theorem are sufficient, but they are not necessary for the sum itself to be integrable. On $[0,1]$, define $f(x)=1$ when $x$ is rational and $f(x)=0$ when $x$ is irrational, and set $g(x)=-f(x)$. Both functions are bounded.

For any partition of $[0,1]$, every subinterval of positive length contains both rational and irrational numbers. Choosing rational tags in every subinterval gives a sum for $f$ equal to

\sum_{i=1}^{n}1\cdot\Delta x_i = \sum_{i=1}^{n}\Delta x_i =1.

Choosing irrational tags in every subinterval instead gives a sum equal to $0$. These two sums occur for partitions of arbitrarily small mesh, and their difference is always $1$. They cannot both approach one common limit. Thus $f$ is not Riemann integrable. The same reasoning, with sums $-1$ and $0$, shows that $g$ is not Riemann integrable.

However, $f(x)+g(x)=0$ at every point. Every tagged sum for $f+g$ is therefore $0$, so $f+g$ is integrable and

\int_0^1(f(x)+g(x))\,dx=0.

This does not contradict the theorem: its assumption that both summands are integrable is a reliable condition guaranteeing the sum rule, not a claim that the sum cannot be integrable otherwise.

Common Pitfall: Using Different Tags

The identity $S(f+g)=S(f)+S(g)$ is exact only when the three sums use the same partition and the same tags. If the tags differ, the three sums are different finite approximations, and their values need not satisfy that identity. In the proof, we are free to use common tags because integrability controls every choice of tags once the mesh is sufficiently small.

A second mistake is to infer integrability of $f+g$ merely from the algebraic identity between sums, without checking whether the sums for $f$ and $g$ have limits. The proof needs both functions to be Riemann integrable: their sums must approach $I_f$ and $I_g$, respectively. Only then does the estimate by $\varepsilon/2+\varepsilon/2$ show that the sums for $f+g$ approach $I_f+I_g$.

Takeaway: Linearity begins with identities that hold for every fixed tagged partition. Integrability ensures the relevant sums converge, and the limit of their exact algebraic combination gives the corresponding integral rule.

Check Your Understanding

Use the tagged-sum identities and integrability conditions to answer these questions.

In the scalar-multiple proof, why is the error bound for $f$ chosen to be $\varepsilon/|c|$ when $c\neq 0$?
Why must the sum-rule proof use the same tags for $f$, $g$, and $f+g$?
What hypotheses guarantee that $f+g$ is Riemann integrable on $[a,b]$?
State the integral of $\alpha f+\beta g$ when $f$ and $g$ are Riemann integrable.
How can $f+g$ be integrable in the cancellation example even though neither $f$ nor $g$ is integrable?

Deriving Linearity From Riemann Sums

What You'll Learn

From Riemann Sums to Linearity

Tagged Sums and Integrability

The Exact Identities Behind the Rules

Why the Rules Hold