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Part CCSIR NET December 2025monotonic-functions-are-always-riemann-integrable-jumps-and-all-integrability-says-nothing-about-continuity

Monotonic functions are always riemann integrable jumps and all integrability says nothing about continuity

Let f:[0,1]→[0,1] be a monotonically increasing function, that is, a≤b implies f(a)≤f(b). For any α∈(0,1)\alpha\in(0,1), let LαL_\alpha⁺=limx→α+f(x)lim_{x\to\alpha^{+}}f(x) and LαL_\alpha⁻=limx→α−f(x)lim_{x\to\alpha^{-}}f(x) denote the right hand and left hand limits respectively, provided they exist. For α∈(0,1)\alpha\in(0,1), if LαL_\alpha⁺ and LαL_\alpha⁻ exist, define Uα=(LαU_\alpha=(L_\alpha⁻,LαL_\alpha⁺) if LαL_\alpha⁻<Lα<L_\alpha⁺, and ∅ if LαL_\alpha⁺≤Lα\le{}L_\alpha⁻. Which of the following statements are true?

  1. A.LαL_\alpha⁺ and LαL_\alpha⁻ exist for every α∈(0,1)\alpha\in(0,1).✓ Correct
  2. B.If f is surjective, then f is continuous.✓ Correct
  3. C.If f is Riemann integrable, then f is continuous.
  4. D.If the left and right hand limits exist at α,β∈(0,1),α≠β\alpha,\beta\in(0,1), \alpha\ne\beta, then Uα∩Uβ=∅U_\alpha\cap{}U_\beta=\emptyset.✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

Related counterexample: |f| Riemann integrable ⇒ f Riemann integrable

More on this topic

The chapter behind this: Riemann integrability — the criterion and the zoo — free to read

From Integration › Riemann integration and criteria

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