Part CCSIR NET December 2025f-maps-into-0-infinity-not-all-of-r-so-the-running-integral-is-always-non-decreasing

F maps into 0 infinity not all of r so the running integral is always non decreasing

Let F be the set of all functions $f : [0,\infty) \to [0,\infty)$ that are Riemann integrable on [0,t] for all $t\in[0,\infty)$ . Which of the following statements are necessarily true?

A.If f∈F is a continuous function and f(n)=0 for all positive integers n, then $\lim(x\to\infty)\int_{0}$ ˣf(t)dt exists in $\mathbb{R}$ .
B.If f∈F is uniformly continuous and $\lim(x\to\infty)\int_{0}$ ˣf(t)dt exists in $\mathbb{R}$ , then $\lim(x\to\infty)f(x)=0$ .✓
C.There exists a function f∈F such that $\lim(x\to\infty)\int_{0}$ ˣf(t)dt exists in $\mathbb{R}$ but $\lim(x\to\infty)f(x)\ne0$ .✓
D.If f∈F is Lebesgue measurable and $\lim(x\to\infty)\int_{0}$ ˣf(t)dt=0, then f=0 almost everywhere on $[0,\infty)$ .✓

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.

See pricing

50 are analysed free — try those first.

Related counterexample: If ∫₀^∞ f converges then f(x) → 0

F maps into 0 infinity not all of r so the running integral is always non decreasing

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