Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / D / D3

False. Take $k(x)=1/(1+|x|)$. This is bounded and measurable on $\mathbb{R}$, and tends to zero as $x\to \infty$, butThus decay to zero alone does not imply Lebesgue integrability.

False. Let $k$ be the indicator of the positive integers. It is measurable, bounded by one and zero almost everywhere, so its Lebesgue integral on $(0,\infty )$ is zero. Yet $k(n)=1$ for every positive integer $n$, so it does not tend to zero at infinity. Lebesgue integrability is unchanged by modifications on a null set, whereas a pointwise limit can change.

Let $H_0={\int }_0^{\pi }|h(u)|du<\infty$. On each period, substitute $x=j\pi +u$ and use periodicity:Summing over $j\ge 0$ gives the convergent geometric boundMeasurability is preserved under multiplication by the exponential, so this proves Lebesgue integrability.

Use the nonnegative parts $h^{+}=\max (h,0)$ and $h^{-}=\max (-h,0)$, so $h=h^{+}-h^{-}$. The zero-mean assumption gives $H:={\int }_0^{\pi }{h}^{+}={\int }_0^{\pi }{h}^{-}<\infty$. On the $k$-th period, each of the positive and negative weighted integrals lies between $H/((k+1)\pi )$ and $H/(k\pi )$. Their difference therefore satisfies the stronger estimateThe sum of the absolute values of these block integrals converges, proving the existence of the finite limit of the integrals from $\pi$ to $n\pi$.

For genuine Lebesgue integrability we instead need the integral of the absolute value. Periodicity givesIf $H>0$, summation diverges by the harmonic series. If $H=0$, then $h=0$ almost everywhere and the function is integrable. ThusThe finite signed block limit therefore need not be a Lebesgue integral.