Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 27H (Probability and Measure)

The monotone convergence theorem states that if $f_n$ are nonnegative measurable functions with $f_n(x)\uparrow f(x)$ almost everywhere, thenwhere either side may be infinite.

The integrals increase and are bounded above by $\int f$, so let their limit be $L\le \int f$. Let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. The setsincrease and cover $\{s>0\}$ up to a null set. Hence continuity of measure from below givesThus $L\ge c\int s$. Taking the supremum over all simple $s\le f$ and then letting $c\uparrow 1$ gives $L\ge \int f$. Therefore equality holds.

Since ${\theta }_n$ is even,For $\xi \ne 0$, integration by parts givesAt $\xi =0$ the continuous limiting value is the area of the triangle,In particular, ${\hat{\theta }}_n\ge 0$.

Regard the same triangular functions as cutoffs in the frequency variable. They satisfySince $\hat{f}\ge 0$, monotone convergence givesFubini's theorem and evenness of ${\theta }_n$ givePart (b) shows that ${\hat{\theta }}_n\ge 0$. Moreover, Fourier inversion at zero yieldsConsequentlyPassing to the limit provesThus one may take the universal constant $\alpha =2\pi$.