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

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.

The monotone convergence theorem states that for measurable functions $0\le f_1\le f_2\le \cdots$ converging pointwise to $f$, the limit is measurable andBoth sides may be infinite. Monotonicity and nonnegativity replace any need for an integrable dominating function.

The partial sums $S_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase pointwise to the extended-valued sum $S={\sum }_{k=1}^{\infty }{a}_k$. The monotone convergence theorem and finite linearity giveA nonnegative measurable function with finite integral is finite almost everywhere; if it were infinite on a positive-measure set, its integral would be infinite. Define it arbitrarily on the null exceptional set if a real-valued representative is desired. Hence the series converges almost everywhere to an integrable function and

Substitute $x=e^{-t}$ to getwhere the second substitution $u=(k+1)t$ reduces the integral to the supplied gamma integral.

On $(0,1)$, $-x\log x\ge 0$. Expand $x^{-x}=e^{-x\log x}={\sum }_{j=0}^{\infty }(-x\log x{)}^j/j!$. The partial sums increase, so the monotone convergence theorem permits termwise integration. The integral of the $j$-th term is $1/(j+1{)}^{j+1}$, including $j=0$. Reindexing givesThis is finite, for instance because the summands with $k\ge 2$ are bounded by $2^{-k}$.

Expand $x^x=e^{x\log x}={\sum }_{j=0}^{\infty }(x\log x{)}^j/j!$. The sum of the integrals of the absolute values of its terms is ${\sum }_{k=1}^{\infty }{k}^{-k}<\infty$ by part (c). Absolute summability therefore permits interchange of the series and the integral. Applying the moment formula from part (c),Equivalently, the absolute exponential series is $x^{-x}$, an integrable majorant, so dominated convergence gives the same interchange.