Past exam of the integration course of the University of Oxford / 2017 (A13756W1, A4)

Source paper: released examination archive, member 2016/T/a13756w1.pdf. The printed cover dates the examination to 20 June 2017.

If ${\sum }_n{m}^{\ast }(A_n)=\infty$, the required inequality is automatic. Otherwise each $A_n$ has finite Lebesgue outer measure. Fix $\epsilon >0$. By the defining infimum, for each $n\ge 1$ choose a countable interval cover $\{J_{n,k}:k\ge 1\}$ withThe family indexed by $(n,k)\in {\mathbb{N}}^2$ is countable and covers ${\bigcup }_n{A}_n$. Since all lengths are nonnegative, its total length is at most ${\sum }_n{m}^{\ast }(A_n)+\epsilon$. Taking the infimum over covers of the union and then letting $\epsilon \downarrow 0$ givesThis proves countable subadditivity of Lebesgue outer measure directly from interval covers, without assuming that any $A_n$ is measurable.

If the intervals $J_k$ cover $A$, their translates $J_k+a$ cover $A+a$ and have the same lengths. Every candidate total length in the definition of $m^{\ast }(A)$ is therefore also a candidate for $m^{\ast }(A+a)$. Taking infima gives $m^{\ast }(A+a)\le m^{\ast }(A)$. Apply the same argument to $A+a$ and the translation $-a$ to obtain the reverse inequality. HenceThis translation invariance of Lebesgue outer measure holds for every subset $A$, including non-measurable sets and sets of infinite outer measure.

A subset $E\subseteq \mathbb{R}$ is a Lebesgue measurable set when it satisfies Carathéodory's criterion: for every $T\subseteq \mathbb{R}$,Suppose $m^{\ast }(A)=0$ and let $B\subseteq A$, allowing $B=A$. The monotonicity of Lebesgue outer measure follows directly from covers: every cover of a larger set covers a smaller set. Thus $m^{\ast }(B)=m^{\ast }(T\cap B)=0$, while $m^{\ast }(T\setminus B)\le m^{\ast }(T)$. ConsequentlyThe reverse inequality is countable subadditivity of Lebesgue outer measure applied to $T=(T\cap B)\cup (T\setminus B)$. Equality follows, even if $m^{\ast }(T)=\infty$. Therefore $A$ and every subset $B\subseteq A$ are Lebesgue measurable sets of measure zero. This is the completeness of Lebesgue measure, obtained without presupposing measurability of the outer-null set.

For a Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$, the preimage $f^{-1}(U)$ of every open set $U$ is a Lebesgue measurable set. For a Borel measurable function, each such preimage must instead be a Borel set. Equivalently, both definitions can test all Borel target sets: the collectionis a sigma-algebra, since preimages preserve complements and countable unions. Containing all open sets therefore makes it contain the Borel sigma-algebra.

Now let $U$ be open. Since $h$ is a Borel measurable function, $h^{-1}(U)$ is Borel. Since $g$ is a Lebesgue measurable function, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set. ButThus $h\circ g$ is a Lebesgue measurable function. The intermediate sigma-algebra is Borel; this is why Borel postcomposition preserves Lebesgue measurability.

The dominated convergence theorem says that if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le H$ almost everywhere for a single Lebesgue integrable function $H\ge 0$, then $f$ is integrable andHere, for $x>0$, put $H_0(x)=x^{-1/2}{e}^{-x}$ and $f_n(x)=H_0(x)(1-(-1{)}^n{e}^{-nx})$. The exponential function gives $e^{-nx}\to 0$ at every such $x$, so $f_n\to H_0$. Also $|f_n(x)|\le 2H_0(x)$. This dominator is integrable: on $(0,1)$, $H_0(x)\le x^{-1/2}$, whose integral is two; on $[1,\infty )$, $H_0(x)\le e^{-x}$, whose integral is finite. DCT therefore proves the required limit.

The limiting value can also be evaluated. Substitute $x=t^2$ and use the Gaussian integral:In fact, scaling the second exponential term gives the exact integral $\sqrt{\pi }[1-(-1{)}^n/\sqrt{n+1}]$. The integral error decays as $n^{-1/2}$ despite exponential convergence at each fixed positive $x$; the slow part lies near the endpoint zero.

Let $\varphi :I\to J$ be a $C^1$ diffeomorphism between intervals, which may be unbounded. For a nonnegative Lebesgue measurable function $f$ on $J$, Lebesgue substitution on an interval givesallowing the value infinity. The same identity applied to $|f|$ shows that $f$ is Lebesgue integrable exactly when the transformed integrand is integrable; in that case the displayed formula holds for signed or complex $f$ as well. The absolute value on the derivative makes the formula valid for both increasing and decreasing substitutions. Finite endpoints can be omitted without changing a Lebesgue integral, because individual points are null sets.

Both functions are continuous on the positive domains in question, so they are measurable. For the first function, the decreasing substitution $t=1/x$ and $|dx/dt|=t^{-2}$ giveFor every integer $k\ge 1$,Summing this lower bound gives a divergent harmonic series. Thus $f$ is not Lebesgue integrable on $(0,1)$.

For the second function, $|\sin x|\le x$ on $(0,1)$ and $|\sin x|\le 1$ everywhere giveTherefore $g$ is Lebesgue integrable on $(0,\infty )$. The sine power kernel integrability threshold is $1<p<2$: one endpoint requires $p<2$, and the other requires $p>1$. The signed improper integral for $f$ does converge by the Dirichlet test for improper integrals, but conditional improper convergence does not give a finite integral of $|f|$.

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.

A sufficient theorem is dominated differentiation under the integral sign. Let $t_0$ be interior to a parameter interval, suppose $u(\cdot ,t)$ is measurable for each $t$, and suppose $u(\cdot ,t_0)$ is integrable. Assume $u(x,\cdot )$ is differentiable near $t_0$ for $x$ outside one fixed null set, and that throughout this neighborhoodfor one nonnegative Lebesgue integrable function $H$. Then $F(t)={\int }_{E}u(x,t)dx$ is defined near $t_0$, is differentiable at $t_0$, andIndeed the mean value theorem bounds both $|u(x,t)-u(x,t_0)|$ by $|t-t_0|H(x)$ and the difference quotient by $H(x)$. The first bound proves local integrability; the second lets the dominated convergence theorem pass the limit of difference quotients through the integral.

Fix $t\in \mathbb{R}$. The mean value theorem for the exponential givesConsequentlyThe function is continuous on $(0,1)$ and hence measurable, so $u(\cdot ,t)$ is Lebesgue integrable for every real $t$. In particular, the numerator cancels the apparent $1/x$ singularity at zero: $u(x,t)\to t$ as $x\downarrow 0$. Only the integrable square-root singularity at one remains. At $t=0$ the function is identically zero.

Fix any real $t_0$ and restrict $t$ to $|t-t_0|<1$. The parameter derivative isThis bound is integrable on $(0,1)$ and is uniform throughout the parameter neighborhood. Part (ii) supplied integrability at $t_0$, so dominated differentiation under the integral sign proves differentiability there. As $t_0$ was arbitrary,For $t>0$, $e^{tx}\ge 1$; for $t<0$, $e^{tx}\le 1$. Since ${\int }_0^1(1-x{)}^{-1/2}dx=2$, these comparisons giveThe first two inequalities are actually strict, because the exponential differs strictly from one throughout $(0,1)$. Differentiating once more under the same type of bound gives $F^{\prime \prime }(t)={\int }_0^{1}x{e}^{tx}/\sqrt{1-x}dx>0$. Hence $F$ is strictly convex and lies above its tangent $2t$ at zero. These properties are collected in exponential integral with a square-root endpoint singularity.

On sigma-finite measure spaces, including the real line with Lebesgue measure, Tonelli's theorem says that a nonnegative measurable function $H$ has equal product and iterated integrals:with infinity allowed. Finiteness is not a prerequisite.

Fubini's theorem says that if a measurable signed or complex $H$ is absolutely integrable on the product, then its sections are integrable for almost every value of the other variable, the iterated integral functions are integrable, and the same equalities hold with finite values. For completed Lebesgue measure, exceptional sections need not themselves be measurable; the integrals use the almost-everywhere defined sections. Thus Tonelli can first establish absolute integrability using $|H|$, after which Fubini permits changing the order for $H$.

Because $f$ is continuous, $g(x,y)=e^{-x^2}f(e^{-x^2/2}y)$ is continuous on ${\mathbb{R}}^2$, so $|g|$ is a nonnegative measurable function. Apply Tonelli's theorem before assuming integrability. For each fixed $x$, set $z=e^{-x^2/2}y$, so $dy=e^{x^2/2}dz$. Lebesgue substitution on an interval givesTherefore the Gaussian integral yieldsThis proves that $g$ is Lebesgue integrable on ${\mathbb{R}}^2$. Fubini's theorem now permits changing the order for $g$ itself. The same substitution without absolute values gives the stronger exact identityTaking absolute values and using the triangle inequality for an integral provesThe constant is sharp: any nonzero nonnegative continuous integrable $f$ gives equality. This is a weighted dilation kernel with $a(x)=e^{-x^2}$ and $b(x)=e^{-x^2/2}$. The inner $x$ integral in the requested order is guaranteed to be integrable for almost every $y$, which is all the iterated integral requires.