2.1 Riemann integration and the fundamental theorem

Upper and lower sums

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Theorem: bounded monotone functions are integrable

For nondecreasing $f$, take a uniform partition with $n$ intervals of width $(b-a)/n$. On interval $i$, the infimum and supremum are the endpoint values. Hence

$$U-L=\frac{b-a}{n}\sum_{i=1}^n[f(x_i)-f(x_{i-1})] =\frac{(b-a)(f(b)-f(a))}{n}\longrightarrow0.$$

The criterion proves integrability even if $f$ has jump discontinuities. For a nonincreasing function apply the result to $-f$.

Theorem: finitely many discontinuities are permitted

If $f$ is bounded and continuous except possibly at finitely many points, it is integrable.

Proof. Let $|f|\le M$, taking $M>0$. Surround the exceptional points by finitely many small intervals, with endpoints included in the partition, having total length less than $\varepsilon/(4M)$. At endpoints of $[a,b]$ use one-sided intervals. Their contribution to the upper-lower gap is at most $2M$ times their total length, hence less than $\varepsilon/2$. On the remaining finitely many closed intervals, $f$ is continuous and therefore uniformly continuous. Subdivide each sufficiently finely that its oscillation on every resulting subinterval is less than $\varepsilon/[2(b-a)]$. Their total gap is less than $\varepsilon/2$. Combining all partition points gives gap below $\varepsilon$. If an exceptional interval endpoint would itself be exceptional, move it slightly so all actual exceptional points lie inside the selected intervals or at the domain endpoints. This keeps the remaining closed intervals continuous.

Proof

Why continuity suffices

A continuous function on a compact interval is uniformly continuous. Given $\varepsilon>0$ and $b>a$, choose $\delta$ so that points less than $\delta$ apart have output difference less than $\varepsilon/(2(b-a))$. A partition with mesh below $\delta$ has oscillation on every subinterval at most that bound, hence $U-L<\varepsilon$. The Darboux criterion yields integrability. For $a=b$ the integral is zero by convention.

Worked example

A bounded function that is not integrable

Let $f(x)=1$ for rational $x\in[0,1]$ and $f(x)=0$ for irrational $x$. Both sets are dense, so every nondegenerate subinterval has supremum $1$ and infimum $0$. Every upper sum is $1$ and every lower sum is $0$. The criterion fails: this bounded function is not Riemann integrable.

Fundamental theorem of calculus

Theorem

Integral and derivative

If $f$ is continuous on $[a,b]$, the accumulation function $F(x)=\int_a^x f(t)dt$ is differentiable on $(a,b)$ with $F'(x)=f(x)$. If $G$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'(x)=f(x)$ for $x\in(a,b)$, then $\int_a^b f(x)dx=G(b)-G(a)$.

Proof

Average-value estimate

For small nonzero $h$ with $x+h\in[a,b]$,

$$\frac{F(x+h)-F(x)}h-f(x)=\frac1h\int_x^{x+h}[f(t)-f(x)]dt.$$

Continuity makes the absolute value of the right-hand side arbitrarily small, including for negative $h$ with oriented-integral convention. Thus $F'=f$. Both $G$ and $F$ are continuous on $[a,b]$, and $G-F$ has derivative zero on $(a,b)$, so it is constant by the mean value theorem; evaluating endpoints gives the second conclusion.

Worked example

Integrate a continuous function

For $f(x)=2x+1$ on $[0,2]$, $G(x)=x^2+x$ is an antiderivative. The theorem gives $\int_0^2(2x+1)dx=G(2)-G(0)=6$. The antiderivative shortcut rests on the theorem, not the definition of an integral.

Method comparison

Use upper and lower sums to prove integrability or produce a counterexample. Use uniform continuity to control oscillation for continuous functions. Once integrability and an antiderivative are available, the fundamental theorem evaluates the integral efficiently. Boundedness alone does not imply Riemann integrability.

Practice

1. Compute $\int_0^1 3x^2\,dx$ by the fundamental theorem.
2. Why is every continuous function on $[0,1]$ Riemann integrable?
3. For the rational-indicator function above, find the upper and lower sum of any partition.
4. If $F(x)=\int_0^x\cos t\,dt$, compute $F'(x)$.

Full solutions

1. An antiderivative is $x^3$, giving $1$.
2. Continuity on a compact interval implies uniform continuity, hence partitions with arbitrarily small upper–lower gap exist.
3. Every subinterval contains rational and irrational points, giving $M_i=1,m_i=0$. Hence $U=1,L=0$.
4. Since cosine is continuous, the first fundamental theorem gives $F'(x)=\cos x$.

Synthesis

Darboux sums define an integral through controlled upper and lower approximations. Uniform continuity guarantees integrability of continuous functions, and the fundamental theorem connects accumulated area to differentiation.