Past exam of the options II course of the University of Oxford / 2008 printed variant / D (Integration)

The monotone convergence theorem says that if measurable functions satisfy $0\le f_1\le f_2\le \cdots$ and converge pointwise to $f$, then $f$ is measurable andallowing the common value to be $+\infty$.

No. Take $f_n=1_{[n,n+1]}$. Each is nonnegative and integrable, and each integral is one. At every fixed real point the functions eventually vanish, so the pointwise limit is zero and its integral is zero. Monotonicity or another convergence hypothesis is essential.

The partial sums $S_N={\sum }_{k=0}^N{a}_k$ are measurable, nonnegative and increasing. By the monotone convergence theorem,A nonnegative measurable function with finite integral is finite almost everywhere: if it were infinite on a positive-measure set, every constant multiple of that set's indicator would be a lower bound, forcing an infinite integral. Thus the series converges almost everywhere.

Substitute $t=\log x$, with $dx=e^{t}dt$. The nonnegative integral becomesby scaling and integration by parts. The exponential decays because $\alpha >1$, with zero boundary term at infinity, proving integrability and

For $x>1$, the geometric series givesAll terms are nonnegative, so monotone convergence permits termwise integration. Part (i), with $\alpha =p+k+1>1$, givesThis series is finite for $p>0$, which also proves integrability of the original nonnegative function.

On a product of sigma-finite measure spaces, Tonelli's theorem equates product and iterated integrals for a nonnegative measurable function, allowing infinite values. Fubini's theorem says that an absolutely integrable measurable function has integrable sections almost everywhere, their section integrals are integrable, and both iterated integrals equal the product integral. Absolute integrability may be checked first by applying Tonelli to the absolute value.

For fixed $n>0$, Tonelli and the allowed exponential integral giveThe point $x=0$ is irrelevant to the integral. Fubini therefore permits reversing the signed integrals. Evaluating the elementary inner integral yieldsThe first term has integral $\pi /2$. The absolute value of the remainder is at mostConsequently the improper integral tends to $\pi /2$. The interchange was justified on each finite strip, without asserting absolute integrability on the infinite quadrant.

It is not Lebesgue integrable, because its absolute integral diverges. On $I_k=[k\pi +\pi /6,k\pi +5\pi /6]$, one has $|\sin x|\ge 1/2$ and $x\le (k+1)\pi$. ThusThese disjoint-interval lower bounds have a divergent harmonic sum. The finite improper value from part (i) is therefore conditional convergence, not Lebesgue integrability.