Countability Of Discontinuities For BV Functions

From Bounded Variation to Countability

A function of bounded variation can oscillate and need not be monotone or continuous. Nevertheless, on a compact interval it cannot have an uncountable set of discontinuities. The key is to decompose it into the difference of two nondecreasing functions: each monotone function has at most countably many discontinuities, and the difference can be discontinuous only where at least one component is discontinuous.

Throughout, let \(a<b\) be real numbers and let \(f:[a,b]\to\mathbb{R}\) be real-valued. Continuity at \(a\) and \(b\) means continuity relative to the domain \([a,b]\). For a partition \(P=(a=x_0<x_1<\cdots<x_n=b)\), its variation sum is \(\sum_{i=1}^n|f(x_i)-f(x_{i-1})|\). The total variation \(V_a^b(f)\) is the supremum of these sums, and \(f\) is of bounded variation, or \(f\in BV[a,b]\), if this supremum is finite.

Theorem: If \(f\in BV[a,b]\), then the set of points in \([a,b]\) at which \(f\) is discontinuous is at most countable. This is a pointwise statement; it does not assert that \(f\) is continuous everywhere.

Why Monotone Functions Have Only Countably Many Discontinuities

First consider a finite-valued nondecreasing function \(u:[a,b]\to\mathbb{R}\). For an interior point \(c\in(a,b)\), define its left and right limits by \(L_c=\sup_{a\le x<c}u(x)\) and \(R_c=\inf_{c<x\le b}u(x)\). These are finite because \(u\) is bounded between \(u(a)\) and \(u(b)\). Monotonicity gives \(L_c\le u(c)\le R_c\). The one-sided limits exist, and \(u\) is continuous at \(c\) exactly when \(L_c=R_c\): when they are equal, the inequalities \(u(x)\le L_c\) for \(x<c\) and \(u(x)\ge R_c\) for \(x>c\) give continuity by squeezing; if they differ, the left and right limits differ, so continuity fails.

Proof. For each discontinuity \(c\in(a,b)\), the open interval \((L_c,R_c)\) is nonempty. Choose a rational number \(q_c\) in this interval. If \(c<d\), choose \(y\) with \(c<y<d\). Then \(R_c\le u(y)\le L_d\), so the intervals \((L_c,R_c)\) and \((L_d,R_d)\) are disjoint. Consequently the rationals \(q_c\) chosen for distinct discontinuities are distinct. Since the rational numbers are countable, there are at most countably many interior discontinuities. The two endpoints \(a,b\) add at most two points. Thus the discontinuity set of \(u\) is at most countable. \(\square\)

The proof counts gaps in the range of a monotone function, not merely jumps of a specified size. This matters because discontinuities can have arbitrarily small jumps; the rational number in each gap provides a countable label regardless of its size.

Jordan Decomposition and the Main Proof

For \(x\in[a,b]\), let \(V_a^x(f)\) denote the total variation of \(f\) on \([a,x]\), with \(V_a^a(f)=0\). The variation is additive on adjacent intervals, so for \(x<y\) we have \(V_a^y(f)-V_a^x(f)=V_x^y(f)\). Also, \(V_x^y(f)\ge |f(y)-f(x)|\), since the partition consisting of the endpoints \(x,y\) is among the partitions used to define variation.

Definition (Jordan components): For \(f\in BV[a,b]\), define \(g(x)=\tfrac12\bigl(V_a^x(f)+f(x)-f(a)\bigr)\) and \(h(x)=\tfrac12\bigl(V_a^x(f)-f(x)+f(a)\bigr)\). These are the normalized nondecreasing components of \(f\).

Proof. If \(x<y\), additivity of variation and its lower bound by the endpoint difference give \(V_a^y(f)-V_a^x(f)\ge |f(y)-f(x)|\). Therefore both \(g(y)-g(x)=\tfrac12\bigl(V_a^y(f)-V_a^x(f)+f(y)-f(x)\bigr)\) and \(h(y)-h(x)=\tfrac12\bigl(V_a^y(f)-V_a^x(f)-f(y)+f(x)\bigr)\) are nonnegative. Thus \(g\) and \(h\) are nondecreasing. They are finite-valued because \(f\) has finite variation and is real-valued, and direct subtraction gives \(g(x)-h(x)=f(x)-f(a)\). Hence \(f(x)=f(a)+g(x)-h(x)\) for every \(x\in[a,b]\). \(\square\)

Proof of the theorem: The preceding argument writes \(f=f(a)+g-h\) with \(g\) and \(h\) nondecreasing. By the monotone discontinuity result, each of \(g\) and \(h\) has at most countably many discontinuities. At any point where both are continuous, their difference \(f(a)+g-h\) is continuous, by the algebra of limits of real-valued functions. Thus every discontinuity of \(f\) belongs to the union of the discontinuity sets of \(g\) and \(h\). A union of two countable sets is countable, which proves the claim. \(\square\)

The conclusion is an inclusion of discontinuity sets, not generally an equality. The two components may have discontinuities at the same points, and their changes can cancel in the difference. The examples below illustrate both the theorem and this distinction.

Worked Example: A Finite Step Function

On \([0,1]\), define \(f(x)=0\) for \(0\le x<1/3\), \(f(x)=2\) for \(1/3\le x<2/3\), and \(f(x)=-1\) for \(2/3\le x\le1\). Find its variation and discontinuities, and compute its Jordan components.

The values change from \(0\) to \(2\) at \(1/3\), then from \(2\) to \(-1\) at \(2/3\). Inserting both breakpoints into any partition cannot decrease its variation sum, by the triangle inequality. The refined sum is at most the sum of the two absolute changes, and a partition containing both breakpoints attains that sum. Therefore \(\displaystyle V_0^1(f)=|2-0|+|-1-2|=2+3=5.\) The function is constant in a relative neighborhood of every point other than the two breakpoints, so its discontinuity set is exactly \(\{1/3,2/3\}\).

The variation function is \(0\) on \([0,1/3)\), \(2\) on \([1/3,2/3)\), and \(5\) on \([2/3,1]\). Since \(f(0)=0\), the normalized components are \(g=(V_0^x(f)+f(x))/2\) and \(h=(V_0^x(f)-f(x))/2\). Their values on the three intervals are respectively \(g=0,2,2\) and \(h=0,0,3\). Thus \(f=g-h\), including at both breakpoints. Check: on the first interval \(g-h=0-0=0\); on the second \(g-h=2-0=2\); on the third \(g-h=2-3=-1\). The total variation calculation also agrees with the endpoint changes \(2\) and \(3\), whose sum is \(5\).

Worked Example: Countably Many Jumps with an Accumulation Point

For \(n\ge1\), let \(r_n=1/(n+1)\), and define on \([0,1]\) \(f(x)=\sum_{n=1}^{\infty}2^{-n}\chi_{[r_n,1]}(x)\). Determine its discontinuities and variation.

Every summand is nondecreasing in \(x\), so the sum is nondecreasing. It is finite-valued since \(0\le f(x)\le\sum_{n=1}^{\infty}2^{-n}=1\). At \(r_n\), the terms that contribute are exactly those with index \(k\ge n\), because \(r_k\le r_n\) exactly when \(k\ge n\). Thus \(f(r_n)=\sum_{k=n}^{\infty}2^{-k}=2^{1-n}\). Immediately to the left of \(r_n\), the contributing indices are \(k\ge n+1\), so the left limit is \(\sum_{k=n+1}^{\infty}2^{-k}=2^{-n}\). The value at \(r_n\) exceeds that left limit by \(2^{-n}>0\); hence every \(r_n\) is a discontinuity.

There are no other discontinuities. At a point \(x>0\) that is not any \(r_n\), choose a neighborhood whose points are bounded below by some positive number. Only finitely many thresholds \(r_n\) lie above that lower bound. The corresponding finite sum is constant locally because the neighborhood avoids those thresholds; all remaining indicators are \(1\) throughout the neighborhood, so their sum is also constant there. At \(0\), if \(0\le x<1/(N+1)\), then no term with \(1\le k\le N\) contributes. Hence \(0\le f(x)\le\sum_{k=N+1}^{\infty}2^{-k}=2^{-N}\), which tends to \(0=f(0)\) as \(N\to\infty\). So \(f\) is continuous at \(0\), and its discontinuity set is exactly \(\{1/2,1/3,1/4,\ldots\}\).

For a nondecreasing function, the variation sum over any partition telescopes to \(f(1)-f(0)\). Here \(f(1)=\sum_{n=1}^{\infty}2^{-n}=1\) and \(f(0)=0\), so \(V_0^1(f)=1\). Check: at \(r_1=1/2\), the value is \(2^{0}=1\), the left limit is \(2^{-1}=1/2\), and the jump is \(1/2\). At \(r_2=1/3\), the value is \(2^{-1}=1/2\), the left limit is \(2^{-2}=1/4\), and the jump is \(1/4\). These agree with the formula \(2^{-n}\) for the jump size.

Worked Example: Cancellation Between Jordan Components

Let \(g(x)=h(x)=\chi_{[1/2,1]}(x)\) on \([0,1]\), and set \(f(x)=g(x)-h(x)\). Compare the discontinuity sets and explain why the theorem gives an inclusion rather than equality.

The functions \(g\) and \(h\) are nondecreasing. For each, the value is \(0\) below \(1/2\) and \(1\) at and above \(1/2\), so each is discontinuous exactly at \(1/2\). But for every \(x\in[0,1]\), \(f(x)=0\), so \(f\) is continuous everywhere and has total variation \(0\). Thus \(D(f)=\varnothing\), whereas \(D(g)\cup D(h)=\{1/2\}\), where \(D(u)\) denotes the discontinuity set of \(u\). Check: for \(x<1/2\), \(g(x)-h(x)=0-0=0\), and for \(x\ge1/2\), \(g(x)-h(x)=1-1=0\). The correct conclusion is \(D(f)\subseteq D(g)\cup D(h)\), not equality.

A Pitfall: Countability Depends on Bounded Variation

It is not enough for a function to be bounded in value; bounded variation is a stronger condition. For example, let \(q:[0,1]\to\mathbb{R}\) equal \(1\) at rational points and \(0\) at irrational points. Both rationals and irrationals are dense in \([0,1]\), so in every neighborhood of every point the function takes both values \(0\) and \(1\). Therefore \(q\) is discontinuous everywhere. In particular, its discontinuity set is uncountable.

Moreover, \(q\) does not have bounded variation. For each positive integer \(N\), choose \(2N+1\) points in strictly increasing order, alternating between rational and irrational points; density allows this choice in successive disjoint subintervals of \([0,1]\). At each consecutive pair, the values differ by \(1\), so the variation sum along these points is \(2N\). Since \(N\) can be arbitrarily large, the variation is infinite. This does not contradict the theorem: the hypothesis \(q\in BV[0,1]\) fails.

A second common error is to argue that every discontinuity of \(f\) must occur at a point where both Jordan components are discontinuous. The theorem only needs the weaker and correct implication: if both components are continuous at a point, then their difference is continuous there. The cancellation example shows why replacing the resulting inclusion by equality is invalid.

Check Your Understanding

Use the decomposition and countability arguments above to answer the following questions.

For a nondecreasing \(u:[a,b]\to\mathbb{R}\), why are the open intervals \((L_c,R_c)\) associated with distinct interior discontinuities disjoint?
Verify directly that the functions \(g\) and \(h\) defined from \(V_a^x(f)\) are nondecreasing when \(f\in BV[a,b]\).
For the countable-jump example, compute the value and left limit at \(r_3=1/4\), and state the jump size.
Why does continuity of both Jordan components at \(x\) imply continuity of \(f\) at \(x\)?
What fails in the argument if one assumes only that \(f\) is bounded in value, rather than of bounded variation?