Proof Continuous Functions Are Integrable

From Pointwise Neighborhoods to One Partition

The Integrability of Continuous Functions theorem established that every continuous function on a closed bounded interval is Riemann integrable. Its proof used uniform continuity to control the Darboux gap on a fine partition. There is another way to organize the central step: begin with the separate neighborhoods supplied by continuity at each point, then use compactness to find one positive scale that works across the whole interval.

The issue is that continuity at a point $c$ gives a neighborhood whose size may depend on $c$. It does not immediately give a single radius that works at every point. Compactness supplies the missing bridge. The result below, called the Lebesgue number lemma, says that an open cover of a closed interval has a positive scale such that every sufficiently small subset lies in one member of the cover.

Lemma (Lebesgue Number Lemma for a Closed Interval): Let $[a,b]$ be a closed bounded interval, and let $\mathcal{U}$ be an open cover of $[a,b]$. There is a number $\lambda>0$ such that every nonempty subset $A\subseteq[a,b]$ with $\operatorname{diam}(A)<\lambda$ is contained in some member of $\mathcal{U}$. Here $\operatorname{diam}(A)=\sup\{|x-y|:x,y\in A\}$.

Proof. Suppose no such $\lambda$ exists. Then for every positive integer $n$, there is a nonempty set $A_n\subseteq[a,b]$ with $\operatorname{diam}(A_n)<1/n$ that is not contained in any member of $\mathcal{U}$. Choose $x_n\in A_n$. By the Bolzano–Weierstrass Theorem, the sequence $(x_n)$ has a subsequence $(x_{n_k})$ converging to some $x\in[a,b]$.

Because $\mathcal{U}$ covers $[a,b]$, some $U\in\mathcal{U}$ contains $x$. Since $U$ is open relative to $[a,b]$, there is an $r>0$ such that $$ (x-r,x+r)\cap[a,b]\subseteq U. $$ For sufficiently large $k$, both $|x_{n_k}-x|<r/2$ and $\operatorname{diam}(A_{n_k})<r/2$. For any $y\in A_{n_k}$, the definition of diameter gives $|y-x_{n_k}|\leq\operatorname{diam}(A_{n_k})<r/2$. Therefore $$ |y-x|\leq |y-x_{n_k}|+|x_{n_k}-x|<r. $$ Thus $A_{n_k}\subseteq U$, contradicting its choice. This contradiction proves the lemma. $\square$

Continuity Gives an Open Cover with Small Oscillation

Fix a desired positive bound $\eta$ on changes in function values. At each point $c$, continuity gives a neighborhood in which every value is within $\eta$ of $f(c)$. Any two values in that neighborhood are then within $2\eta$ of one another. These neighborhoods cover the interval, and the Lebesgue number lemma provides a common scale for fitting small partition cells inside them.

Proposition (Uniform Cell-Oscillation Bound from Local Continuity): Let $f:[a,b]\to\mathbb{R}$ be continuous, and let $\eta>0$. There is a $\lambda>0$ such that every subinterval $J\subseteq[a,b]$ with length less than $\lambda$ satisfies $$ \sup_{x\in J} f(x)-\inf_{x\in J} f(x)\leq 2\eta. $$

Proof. For each $c\in[a,b]$, continuity at $c$ gives $r_c>0$ such that $y\in[a,b]$ and $|y-c|<r_c$ imply $|f(y)-f(c)|<\eta$. Define $$ U_c=(c-r_c,c+r_c)\cap[a,b]. $$ The sets $U_c$ form an open cover of $[a,b]$. Apply the Lebesgue Number Lemma to obtain $\lambda>0$. If $J$ is a subinterval of length less than $\lambda$, then $\operatorname{diam}(J)<\lambda$, so $J\subseteq U_c$ for some $c\in[a,b]$.

For any $y,z\in J$, both points belong to $U_c$. Hence $$ |f(y)-f(z)| \leq |f(y)-f(c)|+|f(c)-f(z)| <2\eta. $$ Thus every pair of values on $J$ differs by less than $2\eta$. Taking the supremum over one value and the infimum over the other gives $$ \sup_{x\in J} f(x)-\inf_{x\in J} f(x)\leq 2\eta. $$ This proves the proposition. $\square$

The conclusion is deliberately stated with $2\eta$, rather than $\eta$: each value is compared to the same center value $f(c)$, so the triangle inequality contributes two errors. Choosing a smaller local tolerance will make the final oscillation bound as small as needed.

Using the Cell Bound to Control the Darboux Gap

We now apply the proposition to partitions. The following proof gives the compactness-based route from pointwise continuity to the Darboux Criterion. The Extreme Value Theorem ensures boundedness, so all upper and lower sums are defined.

Theorem (Darboux-Gap Certificate from Pointwise Continuity): Let $f:[a,b]\to\mathbb{R}$ be continuous, where $a<b$. For every $\varepsilon>0$, there is a partition $P$ of $[a,b]$ such that $$ U(f,P)-L(f,P)<\varepsilon. $$

Proof. By the Extreme Value Theorem, $f$ is bounded on $[a,b]$. Set $$ \eta=\frac{\varepsilon}{4(b-a)}. $$ This is positive because $a<b$. Apply the Uniform Cell-Oscillation Bound from Local Continuity with this $\eta$, obtaining $\lambda>0$. By the proposition that uniform partitions have arbitrarily small mesh, choose a partition $P=\{x_0,\ldots,x_n\}$ with $\|P\|<\lambda$.

For each cell $J_i=[x_{i-1},x_i]$, its length is at most $\|P\|<\lambda$. The proposition therefore bounds its oscillation by $2\eta$. By the Oscillation Formula for the Darboux Gap,

\begin{aligned} U(f,P)-L(f,P) &=\sum_{i=1}^{n}(x_i-x_{i-1}) \left(\sup_{J_i}f-\inf_{J_i}f\right)\\ &\leq\sum_{i=1}^{n}(x_i-x_{i-1})\,2\eta\\ &=2\eta(b-a)\\ &=\frac{\varepsilon}{2} <\varepsilon. \end{aligned}

The function is bounded, and for every $\varepsilon>0$ we have constructed a partition with Darboux gap less than $\varepsilon$. The Darboux Criterion proves that $f$ is Riemann integrable on $[a,b]$. $\square$

Worked Examples: Applying the Local-to-Global Argument

Worked Example: The Cubic Function on a Bounded Interval

Let $f(x)=x^3$ on $[0,2]$. At any $c\in[0,2]$, if $y\in[0,2]$, then

|y^3-c^3| =|y-c||y^2+yc+c^2| \leq 12|y-c|,

because $y^2\leq4$, $yc\leq4$, and $c^2\leq4$. Given a local tolerance $\eta>0$, take $r_c=\eta/12$. Then $|y-c|<r_c$ implies $|f(y)-f(c)|<\eta$. These point-centered neighborhoods cover $[0,2]$. The Lebesgue number lemma gives a $\lambda>0$ such that any cell of length less than $\lambda$ lies inside one of these neighborhoods. Its oscillation is at most $2\eta$.

For a requested gap tolerance $\varepsilon>0$, take $\eta=\varepsilon/8$, since $b-a=2$. A partition with mesh below the resulting $\lambda$ satisfies $$ U(f,P)-L(f,P)\leq 2\eta\cdot2=\frac{\varepsilon}{2}<\varepsilon. $$ The estimate does not require locating the largest value of $r_c$; compactness supplies a common scale from the entire cover.

Worked Example: A Trigonometric Function on a Long Interval

Consider $g(x)=\sin(x^2)$ on $[0,3]$. For $x,y\in[0,3]$, the mean value theorem applied to sine gives $|\sin u-\sin v|\leq|u-v|$. Also, $$ |x^2-y^2|=|x-y||x+y|\leq6|x-y|, $$ since $x+y\leq6$. Therefore $$ |g(x)-g(y)|\leq6|x-y|. $$

Fix $c\in[0,3]$ and choose $r_c=\eta/6$. Whenever $|y-c|<r_c$, the inequality just established gives $|g(y)-g(c)|<\eta$. The neighborhoods $(c-r_c,c+r_c)\cap[0,3]$ cover the interval. By the Lebesgue number lemma, some $\lambda>0$ works for every subset of diameter below $\lambda$. Thus every partition cell of length below $\lambda$ has oscillation at most $2\eta$.

For a gap target $\varepsilon$, choose $\eta=\varepsilon/12$, because the interval length is $3$. Then any partition of mesh below $\lambda$ satisfies $$ U(g,P)-L(g,P)\leq 2\eta\cdot3=\frac{\varepsilon}{2}<\varepsilon. $$ Although this example admits a direct global change bound, the construction follows the same local-neighborhood and compactness steps as the general proof.

Worked Example: The Cube-Root Function at an Endpoint

Let $h(x)=x^{1/3}$ on $[0,1]$. The endpoint $0$ can look troublesome because the derivative of $x^{1/3}$ is not bounded near $0$. Continuity, however, is enough. For $u,v\geq0$, suppose $u\geq v$. Writing $A=u^{1/3}$ and $B=v^{1/3}$, we have $A\geq B$ and $$ u-v=A^3-B^3=(A-B)(A^2+AB+B^2)\geq(A-B)^3, $$ because $A^2+AB+B^2\geq(A-B)^2$. Hence $$ |u^{1/3}-v^{1/3}|\leq|u-v|^{1/3}. $$ If $v\geq u$, exchanging their roles gives the same inequality.

For each $c\in[0,1]$, the choice $r_c=\eta^3$ ensures that $|y-c|<r_c$ implies $$ |h(y)-h(c)|\leq|y-c|^{1/3}<\eta. $$ The Lebesgue number lemma applied to these neighborhoods gives a positive cell-length bound $\lambda$. Taking $\eta=\varepsilon/4$, a partition with mesh below $\lambda$ has gap at most $$ 2\eta(1-0)=\frac{\varepsilon}{2}<\varepsilon. $$ The argument remains valid at $c=0$; no derivative or Lipschitz bound is needed.

Why Compactness Matters Here

The local radii in the proof can vary substantially. In the cube-root example, continuity at $0$ naturally gives a radius proportional to $\eta^3$, while other points may allow much larger neighborhoods. It would be invalid to assume, without proof, that the infimum of all these radii is positive. A family of positive numbers can have infimum zero. The Lebesgue number lemma avoids that mistake: it uses the cover as a whole, rather than taking the infimum of the radii assigned to its centers.

The proof also shows the distinct roles of its hypotheses. Continuity supplies the neighborhoods with controlled changes in function values. Compactness supplies a single scale for small cells. Boundedness, obtained here from the Extreme Value Theorem, allows the Darboux sums to be formed. Once the oscillation on each cell is controlled, the Oscillation Formula for the Darboux Gap converts that local control into a global estimate by summing the cell lengths.

Choose a local tolerance.
Make it small enough that twice the tolerance, multiplied by the interval length, is below the desired Darboux-gap bound.

Use continuity at each point.
Build an open neighborhood on which values stay close to the value at its center.

Apply compactness.
Use a Lebesgue number so that every sufficiently short cell lies inside one of the neighborhoods.

Sum the oscillations.
Apply the Oscillation Formula for the Darboux Gap and then the Darboux Criterion.

Check Your Understanding

Use the local-neighborhood and compactness arguments to answer the following questions.

In the Lebesgue number lemma, why does convergence of $x_{n_k}$ to a point $x$ lead to a contradiction?
If every partition cell has oscillation at most $2\eta$, what upper bound follows for the Darboux gap on $[a,b]$?
Why is it not enough simply to take the infimum of the continuity radii $r_c$?
For $x^3$ on $[0,2]$, verify the bound $|y^3-c^3|\leq12|y-c|$.
Which step in the argument uses compactness, and which step uses the Darboux Criterion?