False. Take the bounded measurable function

k (x) = {(1 + x)^{- 1}, 0, x > 0, x \leq 0.

It tends to zero at positive infinity, but

\int_{0}^{R} ∣ k (x) ∣ d x = lo g (1 + R) ⟶ \infty.

Thus decay to zero alone does not imply Lebesgue integrability.

ii

Solution

False. Let

k = 1_{⋃_{n \geq 1} [n, n + 2^{- n - 2}]} .

This is a bounded measurable function, and the intervals are disjoint, so

\int_{0}^{\infty} ∣ k (x) ∣ d x = \sum_{n \geq 1} 2^{- n - 2} < \infty.

Nevertheless

k (n) = 1

for every positive integer

n

, while

k (n + 1/2) = 0

. Hence it has no limit at infinity. Narrow spikes separate Lebesgue integrability from pointwise decay.

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