Past exam of the integration course of the University of Oxford / 2023 / 1

Writing $\mathcal{L}$ for the collection of Lebesgue measurable sets, saying that $\mathcal{L}$ is a sigma-algebra means that $\mathbb{R}\in \mathcal{L}$, that $\mathbb{R}\setminus A\in \mathcal{L}$ whenever $A\in \mathcal{L}$, and that ${\bigcup }_{n=1}^{\infty }{A}_n\in \mathcal{L}$ whenever every $A_n\in \mathcal{L}$. These properties also imply closure under countable intersections and differences.

A real-valued function $f$ is a Lebesgue measurable function precisely when the inverse image of every Borel set is in $\mathcal{L}$. Equivalently, it suffices to test the half-lines:The equivalence follows because these half-lines generate the Borel sigma-algebra, and inverse images preserve complements and countable unions.

The density of the rational numbers gives a countable way to separate two unequal values:Indeed, $f(x)<g(x)$ holds exactly when some rational number lies strictly between them. Every set inside the union is a Lebesgue measurable set, by the definition of a measurable function and the closure properties of a sigma-algebra. Since $\mathbb{Q}$ is countable, the union is measurable.

First suppose $m(E)<\infty$. The interval-cover definition of Lebesgue outer measure, together with $m^{\ast }(E)=m(E)$, provides bounded open intervals $I_j$ covering $E$ such that ${\sum }_j|I_j|<m(E)+\epsilon$. If the definition uses intervals that are not open, enlarge each by an amount whose total is smaller than the available error. For the open set $U={\bigcup }_j{I}_j$, countable subadditivity and additivity of Lebesgue measure giveThis subtraction is valid because both measures are finite.

For arbitrary $E$, including $m(E)=\infty$, set $E_n=E\cap [n,n+1)$ for $n\in \mathbb{Z}$, and choose positive weights $w_n=2^{-|n|}/3$, whose sum is one. Each $E_n$ has finite Lebesgue measure, so the preceding argument gives an open set $U_n\supseteq E_n$ with $m(U_n\setminus E_n)<\epsilon w_n$. Then $U={\bigcup }_{n\in \mathbb{Z}}{U}_n$ is open, contains $E$, andThus the required form of regularity of Lebesgue measure does not require $E$ itself to have finite measure.

The exact criterion isNecessity follows already from one approximation with error less than one: a compact set $K\subseteq \mathbb{R}$ is bounded, hence has finite Lebesgue measure, so $m(E)=m(K)+m(E\setminus K)<\infty$.

Conversely, suppose $m(E)<\infty$. By continuity from above of a measure, the decreasing sets $E\setminus [-N,N]$ have measures tending to zero. Choose $N\ge 1$ so that $m(E\setminus [-N,N])<\epsilon /2$. Put $B=[-N,N]\setminus E$. By part (b)(i), there is an open set $V\supseteq B$ with $m(V\setminus B)<\epsilon /2$. DefineIt is closed and bounded, hence a compact set by the Heine-Borel theorem, and $K\subseteq E$. The discarded portion satisfiesThe second term is at most $m(V\setminus B)$. This proves the claimed inner regularity of Lebesgue measure with an arbitrarily small absolute error.

False. Take $E=\{0\}$ and any non-measurable set $A\subseteq \mathbb{R}$. The vertical line $\{0\}\times \mathbb{R}$ is a planar Lebesgue null set: each segment $\{0\}\times [-N,N]$ can be enclosed in rectangles of arbitrarily small area, and the line is a countable union of these segments. Consequently its subset $E\times A$ is a Lebesgue measurable set of planar measure zero, by Completeness of Lebesgue measure.

A nonempty factor can therefore have too little Lebesgue measure to transmit the other factor's failure of measurability. Its exceptional vertical section at $x=0$ is consistent with Lebesgue measurable sections being guaranteed only almost everywhere.

True. Suppose instead that $B=A_1\times A_2$ were a Lebesgue measurable set in the plane. The section consequence of Fubini's theorem says that there is a Lebesgue null set $N\subseteq \mathbb{R}$ such thatis a Lebesgue measurable set for every $x\notin N$. This version applies to completed Lebesgue measure; it can be obtained on bounded rectangles and then extended by a countable exhaustion, so no finite-measure assumption on $B$ is needed.

For every $x\in A_1$ the section is exactly $A_2$, which is not measurable. Thus $A_1\subseteq N$. By Completeness of Lebesgue measure, every subset of $N$ is measurable, contradicting the hypothesis on $A_1$. Therefore

This requires an argument using Lebesgue outer measure, since $E$ need not be a Lebesgue measurable set. Some bounded portion $F=E\cap [-N,N]$ has $a=m^{\ast }(F)>0$: otherwise countable subadditivity would give $m^{\ast }(E)=0$. Also $a\le 2N<\infty$.

Suppose, for a contradiction, that every nonempty bounded open interval $I$ satisfies $m^{\ast }(E\cap I)\le \lambda |I|$. Because $a/\lambda >a$, the definition of Lebesgue outer measure gives a countable cover $F\subseteq {\bigcup }_j{I}_j$ by nonempty bounded open intervals withUsing countable subadditivity again,a contradiction. Hence at least one interval in such a cover satisfiesThis high-density interval for outer measure applies even to nonmeasurable sets; the argument does not require the Lebesgue density theorem.