Past exam of the options II course of the University of Oxford / 2004 / D (Integration)

The monotone convergence theorem says that if $0\le f_j\uparrow f$ almost everywhere on a measure space, then $\int f_j\uparrow \int f$, allowing infinity. The dominated convergence theorem says that if $g_j\to g$ almost everywhere and $|g_j|\le H$ for an integrable $H$, then $g$ is integrable and $\int g_j\to \int g$.

For $a\ne 0$, the truncated functions $f_a{1}_{[-j,j]}$ increase to $f_a$. Each has the elementary integral $2\arctan (|a|j)/|a|$. Thus monotone convergence givesFor fixed $x$, eventually $|x|\le n\pi /2$ and $n\sin (x/n)\to x$, so $g_n(x)\to 1/(1+x^2)$. Inside the support the supplied inequality yields $n^2{\sin }^2(x/n)\ge 4x^2/{\pi }^2$. Outside the support $g_n=0$. Consequently the global bound isand this majorant is integrable by the already proved formula with $a=2/\pi$. Dominated convergence now gives $\boxed{\int g_n\to \int (1+x^2{)}^{-1}dx=\pi }$. The majorant is independent of $n$, including on the moving support.

Lebesgue measurability of $g$ means that $g^{-1}(U)$ is a Lebesgue measurable subset of $\mathbb{R}$ for every open subset $U\subseteq \mathbb{R}$. Equivalently one can check all sets $\{x:g(x)>t\}$ with real $t$.

There are two conventions for a step function on the whole real line. Under the bounded-support integration convention it is a finite linear combination of indicators of bounded intervals; under the unrestricted piecewise-constant convention nonzero constant tails are allowed. The conclusions below state both conventions explicitly, since only the first assertion depends on this choice.

Under the bounded-support convention, false in general, true if $f(0)=0$. For the first claim, take $g=0$ and $f\equiv 1$: the composition is one on all of $\mathbb{R}$, and cannot have bounded support. For the second, refine the finitely many defining intervals into disjoint intervals on which $g$ is constant. The composition is constant on each of them and vanishes outside their bounded union because $f(0)=0$. Hence it is another step function.

Under the unrestricted piecewise-constant convention, true in both cases: composition with any function preserves constancy on the same partition, including its unbounded cells. This also applies to the broader measurable-partition convention. Continuity of $f$ is not needed for this particular assertion.

False in both cases. Without the zero-at-zero condition, $g=0$ and $f\equiv 1$ already give an integrable input and a non-integrable composition on $\mathbb{R}$.

Even imposing $f(0)=0$ does not control how fast $f(t)$ approaches zero. Take $g(x)=(1+x^2{)}^{-1}$ and the continuous function $f(t)=\sqrt{|t|}$. The input has integral $\pi$, buthas a divergent integral, because its tails are comparable to $1/|x|$. Thus continuity at zero is weaker than the linear growth estimate that would preserve integrability.

True in both cases. For every open $U$, continuity of $f$ makes $f^{-1}(U)$ open. Thenis Lebesgue measurable by measurability of $g$. The value $f(0)$ has no bearing on this preimage argument.