Past exam of the options II course of the University of Oxford / 2011 / D / D4

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.