Darboux criteria and integrable discontinuous functions

Proof

Prerequisite proofs used below

Continuous real functions on a closed interval are uniformly continuous

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Let $a\le b$ be finite real numbers and let $f:[a,b]\to\mathbb R$ be continuous relative to this interval. Uniform continuity means that for every $\varepsilon>0$ there is $\delta>0$ such that all $x,y\in[a,b]$ with $|x-y|<\delta$ satisfy $|f(x)-f(y)|<\varepsilon$. Fix $\varepsilon>0$. For each $c\in[a,b]$, relative continuity gives $r_c>0$ such that $|f(x)-f(c)|<\varepsilon/2$ for interval points with $|x-c|<r_c$. The intervals $(c-r_c/2,c+r_c/2)$ cover the compact interval. Choose a finite subcover with centres $c_1,\ldots,c_k$, and set $\delta=\min_i r_{c_i}/2>0$. The cover is nonempty since $[a,b]$ is nonempty. Given $x,y$ as above, choose $i$ with $|x-c_i|<r_{c_i}/2$. Then $|y-c_i|<\delta+r_{c_i}/2\le r_{c_i}$, so the triangle inequality gives $|f(x)-f(y)|<\varepsilon/2+\varepsilon/2=\varepsilon$. The same $\delta$ works for all points. This includes a singleton interval and uses the actual finite-cover compactness theorem, rather than citing an unproved scalar Heine–Cantor assertion.

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.

Worked examples and stability

A single jump. On $[0,1]$, let $f=0$ for $x<1/2$ and $f=1$ for $x\ge1/2$. It is monotone and integrable, with integral $1/2$. A partition isolating $1/2$ in an interval of arbitrarily small length makes the only nonzero oscillation contribution arbitrarily small. The value at the jump does not alter the integral.

A bounded nonintegrable function. The rational indicator is one on rational inputs and zero on irrational inputs. Both are dense, so every nondegenerate interval has oscillation one. Every partition on $[0,1]$ has $U-L=1$, violating the criterion.

If $f$ is integrable, so is $|f|$, because $||f(x)|-|f(y)||\le|f(x)-f(y)|$ makes its interval oscillation no larger. If $f,g$ are bounded integrable, their product is integrable: with bounds $M_f,M_g$, the oscillation of $fg$ is at most $M_f$ times that of $g$ plus $M_g$ times that of $f$. Choose partitions controlling both gaps and take their common refinement.

Graded practice

1. Foundation. Estimate $U-L$ for $f(x)=x^2$ on $[0,2]$ with $n$ equal subintervals.
2. Core. Compute the integral of $f(x)=\lfloor x\rfloor$ on $[0,3]$ and justify integrability.
3. Core. Prove that changing finitely many values of an integrable bounded function preserves its integral.
4. Proof. Does integrability of $|f|$ imply integrability of $f$? Supply a bounded counterexample.

Full solutions

1. The monotone formula gives $(2/n)(f(2)-f(0))=8/n$. Thus $n>8/\varepsilon$ guarantees gap below $\varepsilon$.
2. The function is monotone and bounded. It equals zero, one and two on the interiors of the three unit intervals, so the integral is $0+1+2=3$. Endpoint values at integers do not contribute a positive-length interval.
3. The difference $h$ is zero except at finitely many points and bounded by some $C$. Cover those points by intervals of total length as small as desired and include endpoints in a partition. Upper and lower sums of $h$ then have absolute magnitudes at most $C$ times that total length; their common integral is zero. Thus adding $h$ preserves both integrability and the integral value.
4. Let $f=1$ at rational points of $[0,1]$ and $f=-1$ at irrational points. Then $|f|=1$ is integrable. But every upper sum for $f$ is one and every lower sum is minus one, so $f$ is not integrable. The one-way implication cannot be reversed.

Summary and term definitions

The oscillation on an interval is supremum minus infimum. Refinement adds partition points and improves the upper-lower estimates. Riemann integrability requires boundedness and arbitrarily small total weighted oscillation, not continuity at every point. The criterion supplies direct proofs for jumps and stability operations, while dense oscillation can defeat every possible partition.