Past exam of the options II course of the University of Oxford / 2011 / D / D3

False. Take the bounded measurable functionIt tends to zero at positive infinity, butThus decay to zero alone does not imply Lebesgue integrability.

False. LetThis is a bounded measurable function, and the intervals are disjoint, soNevertheless $k(n)=1$ 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.

Set $H={\int }_0^{\pi }|h(t)|dt<\infty$. By periodicity and countable additivity of the Lebesgue integral of a nonnegative function,Thus $f$ is Lebesgue integrable for every $\beta >0$.

Use the nonnegative-part convention $h=h^{+}-h^{-}$, where $h^{+}=\max (h,0)$ and $h^{-}=\max (-h,0)$. The zero-mean assumption givesFor $k\ge 1$, write the integral over the $k$th period as $I_k=P_k-N_k$, whereBoth $P_k$ and $N_k$ lie in the interval $[A/((k+1)\pi ),A/(k\pi )]$. Their difference therefore satisfies the stronger estimateConsequently ${\sum }_{k\ge 1}{I}_k$ converges absolutely as a series of numbers, proving the requested limit of integrals over whole periods. This does not assert absolute integrability of the original function: the cancellation took place inside each period.

Indeed, setting $B={\int }_0^{\pi }|h(t)|dt$, we haveThe harmonic series diverges, so $g$ is Lebesgue integrable only if $B=0$. Conversely $B=0$ means $h=0$ almost everywhere on one period, and hence on all periods, so $g=0$ almost everywhere. ThereforeThis is an example of integrating a mean-zero periodic function divided by x.