Monotonicity Restricts Discontinuities
A monotone function can fail to be continuous, but its values cannot oscillate arbitrarily near a point. For a nondecreasing function, values to the left of a point lie below values to its right. If the function is discontinuous, this ordering leaves a nonempty interval of values that the function skips at that point. The intervals skipped at distinct points do not overlap, so each can be assigned a different rational number. This gives a countability proof without making any assumption that the function is bounded in variation or differentiable.
Throughout, let \(a<b\) be real numbers and let \(f:[a,b]\to\mathbb{R}\) be nondecreasing, meaning \(f(s)\le f(t)\) whenever \(a\le s\le t\le b\). Continuity at an endpoint is relative to the domain \([a,b]\). All statements about \(f\) here are pointwise, not almost everywhere. The degenerate interval \(a=b\) has no discontinuities in its relative topology, so we consider \(a<b\).
The one-sided limits in this definition are actual limits. For example, at an interior point \(x\), put \(L=f(x-)\). Given \(\varepsilon>0\), the definition of supremum gives \(t_0<x\) with \(f(t_0)>L-\varepsilon\). For \(t_0<t<x\), monotonicity yields \(L-\varepsilon<f(t_0)\le f(t)\le L\). Thus \(f(t)\to L\) as \(t\uparrow x\). The infimum argument on the right gives \(f(t)\to f(x+)\) as \(t\downarrow x\). The same arguments, using points in \([a,b]\), apply at the endpoints.
One-Sided Limits Characterize Continuity
Proof. At an interior point, monotonicity implies \(f(x-)\le f(x)\le f(x+)\): every value strictly to the left is at most \(f(x)\), and every value strictly to the right is at least \(f(x)\). If \(f\) is continuous at \(x\), the one-sided limits equal \(f(x)\), which gives the stated equalities. Conversely, suppose all three values equal \(c\). Given \(\varepsilon>0\), the one-sided limit definitions provide \(\delta_1,\delta_2>0\) such that \(x-\delta_1<t<x\) implies \(|f(t)-c|<\varepsilon\), and \(x<t<x+\delta_2\) implies \(|f(t)-c|<\varepsilon\). For \(\delta=\min(\delta_1,\delta_2)\), every \(t\in[a,b]\) with \(0<|t-x|<\delta\) satisfies this bound, and \(f(x)=c\), so \(f\) is continuous at \(x\). At \(a\), continuity relative to \([a,b]\) is exactly the right-hand limit equalling \(f(a)\); at \(b\), it is exactly the left-hand limit equalling \(f(b)\). \(\square\)
In particular, an interior discontinuity is exactly a point where \(f(x-)<f(x+)\). Its jump interval is nonempty, and the function value itself lies somewhere in the closed interval between its one-sided limits. The value \(f(x)\) need not be either endpoint. At the left endpoint a discontinuity means \(f(a)<f(a+)\), and at the right endpoint it means \(f(b-)<f(b)\).
Worked Example: A Jump Whose Value Is Between Its Limits
On \([0,1]\), define \(f(x)=0\) for \(0\le x<1/2\), \(f(1/2)=2/5\), and \(f(x)=1\) for \(1/2<x\le1\). The function is nondecreasing: its values are \(0\), then \(2/5\), then \(1\). At \(x=1/2\), the left limit is \(0\) and the right limit is \(1\), so its jump interval is \((0,1)\). Since \(f(1/2)=2/5\) differs from both one-sided limits, the proposition shows that \(f\) is discontinuous there.
At every other point in \((0,1)\), the function is constant on a neighborhood, so it is continuous. At \(0\), \(f(0)=0=f(0+)\); at \(1\), \(f(1-)=1=f(1)\). Thus the only discontinuity on the whole domain is \(1/2\). Check: \(0<2/5<1\), as required by monotonicity and the one-sided limits.
Disjoint Jump Intervals Give Countability
Proof. For each discontinuity \(x\), take its nonempty jump interval as defined above. We first show that the intervals for different points are disjoint. Let \(x<y\) be points of \([a,b]\) at which jump intervals are defined. Choose \(z\) strictly between \(x\) and \(y\); this is possible because \(x<y\) are real numbers. If \(x\) is an interior point or \(a\), then the upper endpoint of its jump interval is \(f(x+)\). Since \(z>x\), the infimum defining \(f(x+)\) gives \(f(x+)\le f(z)\). If \(y\) is an interior point or \(b\), then its lower endpoint is \(f(y-)\), and \(z<y\) gives \(f(z)\le f(y-)\). Therefore the upper endpoint of the interval at \(x\) is at most the lower endpoint of the interval at \(y\). Their open intervals are disjoint.
Enumerate the rational numbers as \(q_1,q_2,\ldots\). Every nonempty open interval contains a rational number, so for each discontinuity \(x\), there is at least one \(q_j\) in its jump interval. Assign to \(x\) the rational with the smallest such index \(j\). Disjointness ensures that two distinct discontinuities cannot be assigned the same rational. This is an injection from the set of discontinuities into a countable set. Hence the discontinuities are at most countable. \(\square\)
The same conclusion holds for a nonincreasing real-valued function: if \(f\) is nonincreasing, then \(-f\) is nondecreasing and has exactly the same continuity points as \(f\). Thus “monotone” can include either direction.
Worked Example: Locating the Jumps of a Step Function
Define \(f:[0,3]\to\mathbb{R}\) by \(f(x)=\lfloor x\rfloor\) for \(0\le x<3\), and \(f(3)=3\), where \(\lfloor x\rfloor\) is the greatest integer at most \(x\). The function is nondecreasing. At \(x=1\), its left limit is \(0\) and its right limit is \(1\), so the jump interval is \((0,1)\). At \(x=2\), the limits are \(1\) and \(2\), giving \((1,2)\). At the right endpoint \(x=3\), the left limit is \(2\) and the endpoint value is \(3\), giving \((2,3)\).
On each open interval \((0,1)\), \((1,2)\), and \((2,3)\), \(f\) is constant, and \(f(0)=f(0+)=0\). The proposition therefore shows there are no other discontinuities. The three jump intervals are disjoint; for example, the first and second share only the boundary value \(1\), which belongs to neither open interval. Check: the discontinuity set is exactly \(\{1,2,3\}\), and each listed point has a nonempty jump interval.
Worked Example: Countably Many Distinct Jumps
On \([0,1]\), set $$ f(x)=\sum_{n=1}^{\infty}2^{-n}\chi_{[1/n,1]}(x). $$ Each summand is nondecreasing, so the sum is nondecreasing. It is finite everywhere because \(0\le f(x)\le\sum_{n=1}^{\infty}2^{-n}=1\). At \(x=1/n\) for \(n\ge2\), the terms with indices \(k\ge n\) are active, since \(1/k\le1/n\) exactly when \(k\ge n\). Thus $$ f(1/n)=\sum_{k=n}^{\infty}2^{-k}=2^{1-n}, \qquad f((1/n)-)=\sum_{k=n+1}^{\infty}2^{-k}=2^{-n}. $$ The right limit equals \(f(1/n)=2^{1-n}\), so the jump interval is \((2^{-n},2^{1-n})\). At \(x=1\), \(f(1)=\sum_{k=1}^{\infty}2^{-k}=1\), while \(f(1-)=\sum_{k=2}^{\infty}2^{-k}=1/2\); its endpoint jump interval is \((1/2,1)\).
To verify there are no other discontinuities, fix \(x\in(0,1)\) that is not any \(1/n\), and let \(\varepsilon>0\). Choose \(N\) with \(\sum_{k>N}2^{-k}=2^{-N}<\varepsilon\). There is a neighborhood of \(x\) containing none of the finitely many thresholds \(1,1/2,\ldots,1/N\). On that neighborhood the first \(N\) terms are constant. The remaining terms have total value between \(0\) and \(2^{-N}\), so the difference between their values at any two points is at most \(2^{-N}<\varepsilon\). This proves continuity at \(x\). At \(0\), if \(0\le x<1/(N+1)\), only indices \(k>N\) can contribute, so \(0\le f(x)\le2^{-N}\); hence \(f(x)\to0=f(0)\). Therefore the discontinuities are exactly \(1,1/2,1/3,\ldots\). Check: the intervals \((2^{-n},2^{1-n})\) for \(n\ge2\), together with \((1/2,1)\), are disjoint: each pair of consecutive intervals has only a shared endpoint, excluded from both. The jumps are countably infinite, in agreement with the theorem.
A Common Error: Using Only the Function Value
A tempting argument is to assign a rational number from \((f(x-),f(x))\) at every discontinuity. That interval need not be nonempty. For example, on \([0,1]\), let \(g(x)=0\) for \(x\le1/2\) and \(g(x)=1\) for \(x>1/2\). At \(x=1/2\), \(g(x-)=g(x)=0\), so \((g(x-),g(x))=(0,0)\) contains no rational number, even though \(g\) is discontinuous there. The right limit is \(g(x+)=1\), and the correct jump interval \((g(x-),g(x+))=(0,1)\) is nonempty. Use both one-sided limits at interior points, and the one-sided limit together with the endpoint value at a boundary point.
The countability result depends on order: monotonicity ensures both that a discontinuity leaves a genuine interval of values and that intervals from different points do not overlap. It does not say that the discontinuity set must be finite, nor that every jump has the same size. The infinite series example has jumps whose sizes decrease, while the disjoint-interval argument still assigns each discontinuity its own rational number.
Check Your Understanding
Use the one-sided-limit criterion and the jump-interval proof to answer these questions.
- For a nondecreasing function at an interior point \(x\), explain why \(f(x-)\le f(x)\le f(x+)\).
- Can a discontinuity at an interior point have \(f(x)=f(x-)\)? Give an example or explain why not.
- For \(f(x)=\lfloor x\rfloor\) on \([0,2]\), list all discontinuities, including any endpoint discontinuity, and give their jump intervals.
- Why does choosing a rational number in each jump interval define an injection rather than merely a map into the rationals?
- Explain why replacing \(f(x+)\) with \(f(x)\) in the interior jump interval can leave a discontinuity with an empty assigned interval.