Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / D / 3

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.