Past exam of the probability, statistics and options I course of the University of Oxford / 2007 / D (Integration)

The monotone convergence theorem states that if measurable functions $f_n$ are nonnegative and $f_n\uparrow f$ pointwise, then $f$ is measurable andallowing either side to be infinite. The same conclusion holds under almost-everywhere monotone convergence after changing values on a null set. No assumption that the integrals stay bounded is required.

Use the increasing nonnegative measurable functions $f_n(x)=x^{-2}{1}_{[1/n,1]}(x)$. Their pointwise limit is the given function except at irrelevant endpoint choices, and the fundamental theorem of calculus on $[1/n,1]$ givesmonotone convergence therefore gives $\int f=\infty$. Since $f\ge 0$, this is also $\int |f|$, and $f$ is not Lebesgue integrable. Measurability alone does not make a finite integral.

Similarly set $g_n(x)=x^{1/2}{1}_{[1/n,1]}(x)$. These nonnegative measurable functions increase to $g$ almost everywhere. The fundamental theorem of calculus givesBy monotone convergence,The integral is finite and $g\ge 0$, so $g$ is Lebesgue integrable. Identifying Lebesgue and ordinary integrals of continuous functions on compact intervals justifies each truncated calculation.

A dominated differentiation under the integral sign suffices. Suppose $\varphi (\cdot ,t)$ is measurable, $\varphi (\cdot ,t_0)$ is integrable at one parameter value, and for almost every $x$ the map $t\mapsto \varphi (x,t)$ is continuously differentiable on an interval. If $|{\partial }_t\varphi (x,t)|\le H(x)$ throughout that interval for an integrable $H$, then the parameter integral is differentiable, andOne direct sufficient version assumes ${\partial }_t\varphi (x,t)$ is continuous in $t$ almost everywhere and has that common domination; dominated convergence theorem then proves the assertion.

Here $|f(x)e^{itx}|=|f(x)|$, so the complex-valued function $h$ is integrable for every real $t$. For a real increment $u\ne 0$, the elementary bound $|e^{iux}-1|\le |ux|$ givesThe right side is integrable by hypothesis and the integrand converges pointwise to $ixf(x)e^{itx}$. dominated convergence theorem, applied to real and imaginary parts, provesThis supplies the domination explicitly rather than merely differentiating a formal Fourier transform.