Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / D (Integration)

The monotone convergence theorem states that if $f_n$ are measurable, nonnegative functions on a measure space with $f_n\le f_{n+1}$ almost everywhere and $f_n\uparrow f$ almost everywhere, thenThe equality is valid in $[0,\infty ]$, so finiteness is not required. Changes on null sets do not affect the conclusion.

No. Take $f_n=1_{[n,n+1]}$ on the real line. At every fixed point it eventually vanishes, so $f=0$, while $\int f_n=1$ for every $n$. Thus the limit of the integrals is one and the integral of the pointwise limit is zero. Nonnegativity alone cannot replace monotonicity or an integrable domination hypothesis.

Let $S_N={\sum }_{k=0}^N{a}_k$. These are nonnegative measurable functions increasing to the extended-valued sum $S$. Monotone convergence givesA nonnegative function with finite integral cannot be infinite on a set of positive measure. Hence $S(x)<\infty$ almost everywhere, which is exactly almost-everywhere convergence of the nonnegative series. One may redefine the sum arbitrarily on the exceptional null set to obtain an ordinary integrable representative.

Substitute $u=\log x$, giving $dx=e^{u}du$. The nonnegative integral becomesFor $\alpha >1$ an integration by parts gives $1/(\alpha -1{)}^2$; its boundary term vanishes because exponential decay dominates $u$. Hence it is finite, proving integrability and

For $x>1$ the geometric expansion givesEvery summand is nonnegative, so monotone convergence permits integration term by term without a prior integrability assumption. The preceding part applies with $\alpha =p+k+1>1$, yieldingThe series is finite for every fixed $p>0$, by comparison of its tail with $\sum k^{-2}$. This also proves integrability of the original function.

On the product of two sigma-finite measure spaces, Tonelli's theorem states that a nonnegative measurable function has measurable iterated-integral functions andwith values allowed to be infinite. Fubini's theorem states that if $f$ is measurable and ${\int }_{X\times Y}|f|<\infty$, then its sections are integrable almost everywhere, both iterated integrals exist as finite integrals, and the same equality holds, also for real or complex signed functions. Tonelli applied to $|f|$ can first be used to verify the absolute-integrability hypothesis of Fubini.

For fixed $n>0$, Tonelli givesbecause the quotient is bounded near zero. Fubini therefore permits reversing the order for this finite $x$ interval. Direct integration in $x$ yieldsThus the desired integral equals $\pi /2$ minus the integral of the exponential remainder. Its absolute value is at mostConsequently $\boxed{\underset{n\to \infty }{\lim }{\int }_0^n\sin x/xdx=\pi /2}$. The truncation is essential: Fubini is not being applied to an absolutely integrable function over the whole infinite rectangle.

No. On each interval $[k\pi ,(k+1)\pi ]$, for $k\ge 1$,The resulting harmonic lower bound diverges. Therefore $\sin x/x\notin L^1(0,\infty )$. Its convergent improper integral does not make it Lebesgue integrable; its positive and negative parts both have infinite integral.