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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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For an integrable f, the one-sided derivatives of F(x)=∫axf equal the corresponding one-sided limits of f; at a jump they are unequal

Statement

Let a<b, let f:[a,b]→R be Riemann integrable, and put F(x)=∫axf.

  1. If c∈[a,b) and lim⁡x→c+f(x)=L+, then the right derivative exists and F+′(c)=L+.
  2. If c∈(a,b] and lim⁡x→c−f(x)=L−, then the left derivative exists and F−′(c)=L−.

In particular, if both one-sided limits exist at an interior point and are unequal, then F is not differentiable there. The value f(c) itself is irrelevant to both conclusions.

Facts & Assumptions

Given: The integrable f, its integral function F, and the indicated one-sided limits.

[L2]

The right-limit condition says that for every ε>0, ∣f(x)−L+∣<ε throughout a sufficiently short interval to the right of c; the left version is analogous (The left and right limits of f at c, as limits of the restrictions of f to A∩(−∞,c) and A∩(c,∞)).

Proof

technique · epsilon-delta
1.1

Assume the right limit exists and fix ε>0. By [L2], choose δ>0 so that ∣f(x)−L+∣<ε whenever c<x<c+δ within [a,b].

givenL2
1.2

For the left limit, take h<0, rewrite the same quotient using the oriented integral over [c+h,c], and apply [L2] and [L3]; its limit is L−.

givenL1L2L3
2.1

For 0<h<δ with c+h≤b, [L1] and linearity give F(c+h)−F(c)h−L+=1h∫cc+h(f−L+).

step 1.1L1algebra
3.1

By [L3], the absolute value in step 2.1 is at most ε. Hence the right difference quotient tends to L+.

step 1.1step 2.1L3
4.1

At an interior point a two-sided derivative would have to equal both one-sided derivatives, so unequal L+ and L− preclude it. Neither estimate refers to f(c).

step 3.1step 1.2∎

Depends on

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