Past exam of the options II course of the University of Oxford / 2013 / D3

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise for measurable functions, then $h$ is measurable and $\int h={\lim }_n\int h_n$, allowing infinity.

Apply it to the partial sums $S_N={\sum }_{n=0}^N{f}_n$. They are nonnegative, measurable, and increase pointwise to $f$. Finite additivity of the integral and monotone convergence yieldThe assumed finite sum of integrals makes $f$ integrable. This proves the series-integration rule needed below directly from the stated theorem, without assuming an additional interchange theorem.

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

For $x>0$, the geometric series givesEvery summand is nonnegative and has integral $(n+\alpha {)}^{-2}$ by part (a). This numerical series converges by comparison with ${\sum }_{n\ge 1}{n}^{-2}$. The series-integration rule proved above therefore yieldsThe value at the single endpoint zero does not affect this Lebesgue integral.

On $0<x<1$, the geometric expansion gives summands $h_n(x)=x^{n+\beta -1}-x^{n-\beta }$. Their integrals areBoth powers are integrable because $0<\beta <1$. If $\beta <1/2$, each $h_n$ is nonnegative, since its first exponent is smaller. Their integrals form a convergent nonnegative series: the $n=0$ term is finite and subsequent terms are bounded by a constant times $n^{-2}$. The proved series-integration rule applies.

If $\beta >1/2$, apply precisely that rule to $-h_n$, which are nonnegative, and negate the result. At $\beta =1/2$ all summands and the integrand are zero. Thus in every case the integral is absolutely integrable andThe common sign of the summands is the reason no unproved theorem for integrating arbitrary signed series is needed.