Past exam of the options II course of the University of Oxford / 2011 / D (Integration)

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.

The monotone convergence theorem states that if $f_n$ are nonnegative measurable functions on a measure space and $f_n\uparrow f$ almost everywhere, thenEither side is allowed to be $+\infty$; no integrable dominating function is required.

The partial sums $s_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase to an extended-valued function $s$. By the monotone convergence theorem,A nonnegative function with finite Lebesgue integral cannot equal $+\infty$ on a set of positive measure: otherwise its integral would exceed $M$ times that measure for every $M>0$. Thus the series converges to a finite value almost everywhere. Redefine it to be zero on the exceptional null set if a finite-valued representative is desired. The resulting function is Lebesgue integrable, with

Put $x=e^{-t}$. Thenwhere the second substitution is $u=(k+1)t$ and the final integral is the given gamma function value. The formula also holds for $k=0$, with value $1$.

For $0<x<1$, the exponential function expansion givesEvery term is nonnegative. By the monotone convergence theorem, integration can be interchanged with summation, and henceThis identity is one form of Sophomore's dream.

Expand $x^x=e^{x\log x}$. The absolute values of all terms have total integralThe series converges, for instance because $j^{-j}\le 2^{-j}$ for $j\ge 2$. Part (b) applied to the absolute values, followed by subtraction of positive and negative parts, therefore justifies termwise integration. It gives the other Sophomore's dream identity:Equivalently, the partial sums are dominated by $e^{|x\log x|}\le e^{1/e}$, so the dominated convergence theorem gives the same justification.