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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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The jumps of a variation function equal the absolute jumps of the original function

Statement

Let f:[a,b]→R have bounded variation and let Vf(x)=Var⁡[a,x](f). At an interior point c,

Vf(c+)−Vf(c)=∣f(c+)−f(c)∣,Vf(c)−Vf(c−)=∣f(c)−f(c−)∣.

The corresponding one-sided formula holds at either endpoint. In particular, Vf is continuous at every point where f is continuous.

Facts & Assumptions

Given: A bounded-variation function f:[a,b]→R, its variation function Vf, and a point c∈[a,b].

[L1]

Vf(y)−Vf(x)=Var⁡[x,y](f) whenever x≤y (Total variation is additive over adjacent subintervals and decreases under restriction).

[L2]

∣f(y)−f(x)∣≤Var⁡[x,y](f) (Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant).

Proof

technique · direct
1.1

Since Vf is nondecreasing and bounded above by Vf(b), its one-sided limits exist. By [L1] and [L2], Vf(x)−Vf(c)≥∣f(x)−f(c)∣ for x>c; passage to the right limit gives Vf(c+)−Vf(c)≥∣f(c+)−f(c)∣.

L1L2L3L4
1.2

For the reverse inequality, fix x0>c and put xn=c+(x0−c)2−n. Let an=Var⁡[xn+1,xn](f). By repeated additivity, every partial sum of the nonnegative series ∑nan is Var⁡[xN,x0](f) for a suitable N, hence is bounded by Var⁡[c,x0](f). Its tails therefore tend to zero by [L6], while xn↓c by [L7].

L1L6L7
2.1

Given ε>0, take N so large that the series tail from N is below ε and ∣f(y)−f(c+)∣<ε whenever c<y≤xN. For any partition c=t0<t1<⋯<tk=xN, choose m≥N with xm+1<t1≤xm. The part after its first increment is at most [step 1.2, L1, L2, L3, L6] Var⁡[t1,xN](f)≤Var⁡[xm+1,xN](f)=∑n=Nman<ε, while ∣f(t1)−f(c)∣≤∣f(c+)−f(c)∣+ε. Taking the supremum over partitions gives Var⁡[c,xN](f)≤∣f(c+)−f(c)∣+2ε. Restriction gives the same bound for c<x≤xN, and [L2] gives the reverse bound in the limit.

3.1

Thus lim⁡x↓cVar⁡[c,x](f)=∣f(c+)−f(c)∣, and [L1] proves the right-hand formula. Applying steps 1.2–2.1 to the reversed interval proves the left-hand formula. If f is continuous at c, both absolute jumps vanish by [L5], so Vf is continuous there. Endpoint cases use only the available side.

step 1.1step 2.1L1L3L5∎

Depends on

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Sources