Past exam of the integration course of the University of Oxford / 2017 / 1

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.