Past exam of the integration course of the University of Oxford / 2017 / 2 / b

The monotone convergence theorem states that, for nonnegative measurable $f_n$ on a Lebesgue measurable set $E$ with $f_n\uparrow f$ almost everywhere,allowing infinity. Fatou's lemma states that, for nonnegative measurable $f_n$,Now write $C={\sup }_n{\int }_E|g_n|<\infty$. A pointwise limit superior of measurable functions is measurable, and it agrees with $g$ off the exceptional null set. By completeness of Lebesgue measure, assigning any values on that null set preserves measurability. Therefore $g$ is measurable. Since $|g_n|\to |g|$ almost everywhere, Fatou's lemma givesThus $g$ is finite almost everywhere and is a Lebesgue integrable function. This is an almost-everywhere limit under a uniform L1 bound. The argument proves integrability of the limit, not convergence in L1 space.

Let $S_N={\sum }_{n=1}^N|a_n|$. These are nonnegative measurable functions with $S_N\uparrow S={\sum }_{n\ge 1}|a_n|$. The monotone convergence theorem and linearity of the finite sum giveHence $S<\infty$ almost everywhere: if $S=\infty$ on a set of positive measure, the integral would exceed every finite constant. Also each $a_n$ is finite almost everywhere, since its absolute integral is finite. Outside the union of these countably many null sets, all terms are real andThus the signed series converges absolutely almost everywhere, as asserted by summable L1 norms imply almost-everywhere absolute convergence. Removing the common null set avoids any undefined $+\infty -\infty$ in signed partial sums.

Let $A(x)={\sum }_{n\ge 1}{a}_n(x)$ on the full-measure set where part (A) gives absolute convergence, and set $A=0$ on its null complement. After similarly replacing the individual terms by zero on that common null set, the partial sums are finite measurable functions and converge to $A$. Hence $A$ is a Lebesgue measurable function. The triangle inequality givesalmost everywhere. Part (A) proved $S$ integrable, soTherefore the summed function is Lebesgue integrable. Its values on the exceptional null set do not affect the integral. This is the integrability step in integration of a series with summable L1 norms.

Let $A_N={\sum }_{n=1}^N{a}_n$, using the finite representatives from part (B). Then $A_N\to A$ almost everywhere, and $|A_N|\le S={\sum }_n|a_n|$, where $S$ is integrable by part (A). The dominated convergence theorem givesThe numerical series on the right converges absolutely because $|{\int }_E{a}_n|\le {\int }_E|a_n|$ and the sum of these bounds is finite. Moreover,Thus integration of a series with summable L1 norms also yields convergence of the partial sums in L1 norm.

Use the power series of the exponential function:Each term is a Lebesgue integrable function, andThe sum of these nonnegative integrals is at most $2{\sum }_{n\ge 0}1/n!=2e<\infty$. Applying integration of a series with summable L1 norms, just proved in part (ii), therefore justifies exchanging the series and the integral:The absolute-integral bound is the justification; pointwise convergence of the exponential series alone would not suffice.