The Riemann Integral

The integral assigns an area under a graph, and the Riemann construction does it by trapping the area between sums that overestimate and sums that underestimate, calling the function integrable when the two can be brought together. The construction rests on the completeness of the reals through the supremum and on uniform continuity to control the gap, and it culminates in the fundamental theorem of calculus, which makes the integral the inverse of the derivative. This post builds it for bounded functions on a closed interval [1], [2].

#Upper and lower sums

Definition1

A partition of $[a,b]$ is a finite set $a=t_0<t_1<\cdots<t_m=b$ . For a bounded $f$ , the lower and upper sums are

L(f,P)=\sum_{i=1}^m\Big(\inf_{[t_{i-1},t_i]}f\Big)(t_i-t_{i-1}),\qquad U(f,P)=\sum_{i=1}^m\Big(\sup_{[t_{i-1} ,t_i]}f\Big)(t_i-t_{i-1}). \tag{1}

The lower integral is $\underline{\int}f=\sup_P L(f,P)$ and the upper integral is $\overline{\int}f =\inf_P U(f,P)$ . The function is Riemann integrable when these are equal, their common value the integral $\int_a^b f$ .

Refining a partition by inserting points raises lower sums and lowers upper sums, because splitting a subinterval replaces one infimum by two larger ones and one supremum by two smaller ones. Consequently any lower sum is at most any upper sum, since both are comparable to their common refinement, and so $\underline{\int}f\le\overline{\int}f$ always. Integrability is the statement that no gap remains.

#The Riemann criterion

The definition asks a supremum and infimum to coincide. The criterion restates this as a single approximation that is easier to verify.

Theorem2

A bounded $f$ is Riemann integrable on $[a,b]$ if and only if for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$ .

Proof

Suppose the gap can be made small. For any $\varepsilon$ pick $P$ with $U(f,P)-L(f,P)<\varepsilon$ . Then $\overline{\int}f-\underline{\int}f\le U(f,P)-L(f,P)<\varepsilon$ , using $\underline{\int}f\ge L(f,P)$ and $\overline{\int}f\le U(f,P)$ . Since this holds for every $\varepsilon$ , the nonnegative difference $\overline{\int}f-\underline{\int}f$ is zero, so $f$ is integrable. Conversely, suppose $f$ is integrable with $\underline{\int}f=\overline{\int}f=I$ . Given $\varepsilon$ , the supremum defining $\underline{\int}f$ gives a partition $P_1$ with $L(f,P_1)>I-\varepsilon/2$ , and the infimum gives $P_2$ with $U(f,P_2)<I+ \varepsilon/2$ . Their common refinement $P$ satisfies $L(f,P)\ge L(f,P_1)$ and $U(f,P)\le U(f,P_2)$ by the refinement monotonicity, so $U(f,P)-L(f,P)<(I+\varepsilon/2)-(I-\varepsilon/2)=\varepsilon$ .

The criterion reduces integrability to making the total oscillation of $f$ , summed against the subinterval lengths, as small as desired. Continuity makes this automatic.

#Continuous functions are integrable

Theorem3

Every continuous function on $[a,b]$ is Riemann integrable.

Proof

Let $\varepsilon>0$ . By the Heine-Cantor theorem $f$ is uniformly continuous on $[a,b]$ , so there is a $\delta>0$ with $\abs{f(s)-f(t)}<\varepsilon/(b-a)$ whenever $\abs{s-t}<\delta$ . Take any partition $P$ with every subinterval shorter than $\delta$ . On each subinterval the extreme value theorem makes the supremum and infimum of $f$ attained values $f(s_i)$ and $f(u_i)$ at points of the subinterval, so $\abs{s_i-u_i}$ is below the subinterval length and hence below $\delta$ , and the oscillation $\sup-\inf=f(s_i)-f(u_i)< \varepsilon/(b-a)$ . Summing,

U(f,P)-L(f,P)=\sum_{i=1}^m\big(\textstyle\sup_{[t_{i-1},t_i]}f-\inf_{[t_{i-1},t_i]}f\big)(t_i-t_{i-1}) <\frac{\varepsilon}{b-a}\sum_{i=1}^m(t_i-t_{i-1})=\varepsilon. \tag{2}

By Theorem 2, $f$ is integrable.

The integral so constructed is monotone, in the sense that $f\le g$ gives $\int f\le\int g$ , and additive over adjacent intervals, both read off the corresponding statements for the sums, since $\inf$ and $\sup$ are monotone and a partition containing the split point decomposes $L$ and $U$ additively. Linearity needs more, because $\inf$ and $\sup$ are only super- and subadditive, giving $L(f,P)+L(g,P)\le L(f+g,P)$ and $U(f+g,P)\le U(f,P)+U(g,P)$ . Given $\varepsilon>0$ , pick partitions with $U(f,P_f)-L(f,P_f)<\varepsilon/2$ and $U(g,P_g)-L(g,P_g)<\varepsilon/2$ and let $P=P_f\cup P_g$ be their common refinement, which preserves both gaps. Then $U(f+g,P)-L(f+g,P)\le(U(f,P)+U(g,P))-(L(f,P)+L(g,P))<\varepsilon$ , so $f+g$ is integrable by Theorem 2; since both $\int(f+g)$ and $\int f+\int g$ lie in $[L(f,P)+L(g,P),\,U(f,P)+U(g,P)]$ , of width $<\varepsilon$ , letting $\varepsilon\to0$ gives $\int(f+g)=\int f+\int g$ , and $\int(cf)=c\int f$ follows from $\inf(cf)=c\inf f$ for $c\ge0$ and the $\sup$ / $\inf$ swap for $c<0$ . With existence settled for continuous functions, the integral becomes a function of its upper limit, and that function is differentiable.

#The fundamental theorem of calculus

Theorem4

Let $f$ be continuous on $[a,b]$ and $F(x)=\int_a^x f(t)\,dt$ . Then $F$ is differentiable with $F'(x)=f(x)$ .

Proof

Fix $x$ and $h\neq 0$ small. By additivity, $F(x+h)-F(x)=\int_x^{x+h}f(t)\,dt$ under the orientation convention $\int_x^{x+h}=-\int_{x+h}^x$ for $h<0$ , so

\frac{F(x+h)-F(x)}{h}-f(x)=\frac1h\int_x^{x+h}\big(f(t)-f(x)\big)\,dt. \tag{3}

Given $\varepsilon>0$ , continuity at $x$ provides $\delta$ with $\abs{f(t)-f(x)}<\varepsilon$ for $\abs{t-x}<\delta$ . For $\abs h<\delta$ monotonicity applied to $-\varepsilon\le f(t)-f(x)\le\varepsilon$ over the interval of integration bounds the integral by $\varepsilon\abs h$ in absolute value. When $h>0$ this is $\abs{\int_x^{x+h}(f(t)-f(x))\,dt}\le\varepsilon h$ on $[x,x+h]$ ; when $h<0$ the same estimate on $[x+h,x]$ gives $\abs{\int_{x+h}^x(f(t)-f(x))\,dt}\le\varepsilon\abs h$ , and $\frac1h\int_x^{x+h}=\frac1{\abs h}\int_{x+h}^x$ matches it. Either way the right side of Equation (3) has absolute value at most $\frac1{\abs h}\,\varepsilon\,\abs h=\varepsilon$ . Hence the difference quotient tends to $f(x)$ , which is $F'(x)=f(x)$ .

Theorem5

If $f$ is continuous on $[a,b]$ and $G$ is any antiderivative of $f$ , meaning $G'=f$ , then $\int_a^b f=G(b) -G(a)$ .

Proof

Let $F(x)=\int_a^x f$ . By Theorem 4, $F'=f=G'$ , so $(F-G)'=0$ on $[a,b]$ , and a function with zero derivative on an interval is constant by the mean value theorem. Thus $F-G=c$ for a constant $c$ . Evaluating at $a$ , where $F(a)=0$ , gives $c=-G(a)$ , and at $b$ , $\int_a^b f=F(b)=G(b)+c=G(b)-G(a)$ .

The two halves together say differentiation and integration undo each other, the antiderivative computing the integral and the integral recovering the function. This is the computational engine of calculus, and it is also the structural statement that the integral of a continuous function is a continuously differentiable function of its limits. The Riemann integral is enough for continuous integrands and for the Stieltjes and mean-square integrals built on the same partition idea, while integrands with worse discontinuities require the Lebesgue integral, which replaces partitions of the domain by partitions of the range and integrates a far larger class.

[1]

W. Rudin, Principles of Mathematical Analysis, 3rd ed. McGraw-Hill, 1976.

[2]

S. Abbott, Understanding Analysis, 2nd ed. Springer, 2015.

Explore connections

see in the atlas

referenced by (1)

Holomorphic Functions and Cauchy's Theorem

cite

@misc{riemann-integral,
  author = {Zac Kienzle},
  title  = {The Riemann Integral},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/riemann-integral}
}

#Upper and lower sums

Definition1

A partition of $[a,b]$ is a finite set $a=t_0<t_1<\cdots<t_m=b$ . For a bounded $f$ , the lower and upper sums are

L(f,P)=\sum_{i=1}^m\Big(\inf_{[t_{i-1},t_i]}f\Big)(t_i-t_{i-1}),\qquad U(f,P)=\sum_{i=1}^m\Big(\sup_{[t_{i-1} ,t_i]}f\Big)(t_i-t_{i-1}). \tag{1}

#The Riemann criterion

The definition asks a supremum and infimum to coincide. The criterion restates this as a single approximation that is easier to verify.

Theorem2

A bounded $f$ is Riemann integrable on $[a,b]$ if and only if for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$ .

Proof

The criterion reduces integrability to making the total oscillation of $f$ , summed against the subinterval lengths, as small as desired. Continuity makes this automatic.

#Continuous functions are integrable

Theorem3

Every continuous function on $[a,b]$ is Riemann integrable.

Proof

U(f,P)-L(f,P)=\sum_{i=1}^m\big(\textstyle\sup_{[t_{i-1},t_i]}f-\inf_{[t_{i-1},t_i]}f\big)(t_i-t_{i-1}) <\frac{\varepsilon}{b-a}\sum_{i=1}^m(t_i-t_{i-1})=\varepsilon. \tag{2}

By Theorem 2, $f$ is integrable.

#The fundamental theorem of calculus

Theorem4

Let $f$ be continuous on $[a,b]$ and $F(x)=\int_a^x f(t)\,dt$ . Then $F$ is differentiable with $F'(x)=f(x)$ .

Proof

Fix $x$ and $h\neq 0$ small. By additivity, $F(x+h)-F(x)=\int_x^{x+h}f(t)\,dt$ under the orientation convention $\int_x^{x+h}=-\int_{x+h}^x$ for $h<0$ , so

\frac{F(x+h)-F(x)}{h}-f(x)=\frac1h\int_x^{x+h}\big(f(t)-f(x)\big)\,dt. \tag{3}

Theorem5

If $f$ is continuous on $[a,b]$ and $G$ is any antiderivative of $f$ , meaning $G'=f$ , then $\int_a^b f=G(b) -G(a)$ .

Proof

[1]

W. Rudin, Principles of Mathematical Analysis, 3rd ed. McGraw-Hill, 1976.

[2]

S. Abbott, Understanding Analysis, 2nd ed. Springer, 2015.

Explore connections

see in the atlas

referenced by (1)

Holomorphic Functions and Cauchy's Theorem

cite

@misc{riemann-integral,
  author = {Zac Kienzle},
  title  = {The Riemann Integral},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/riemann-integral}
}