Correct answer: (D) $S_{1/\sqrt\pi}$ is measurable.
Explanation
f is a countable sum of measurable indicator functions and is therefore measurable, making every strict superlevel set measurable. In particular $S_{1/\sqrt\pi}$ is measurable, D. Also $S_{1/2}=(0,1]$ is not open, $S_0=(0,\infty)$ is not closed, and $S_{\sqrt5/2}$ is empty and hence measurable.
More Measure and Integration questions
GATE 2026 MA Q13– Let C be the Cantor set. On [0,1], define f(x) = x²⁰²⁶ on C; cos(πx) on [0,1/2]∖C; sin(πx)…
GATE 2024 MA Q14– Evaluate the source limit 0-1 0 e-x/[3x+x2+x+4]dx.
GATE 2024 MA Q38– Which condition suffices for f:[0,1]→ℝ to be Lebesgue measurable?
GATE 2023 MA Q16– Let N⊆ℝ be non-Lebesgue-measurable. P:the irrational points of N form a measurable set.…
GATE 2023 MA Q17– Measurable E k⊆[0,1] have measure 1/k². Define E= n k nE k and F= n k nE k. P:m(E)=0;…
GATE 2023 MA Q47– Let f n:ℝ→ℝ equal n²(x−n) on [n,n+1/n²], and zero elsewhere. Write m for Lebesgue measure.…