2.3 Positive linear functionalsCBAFDF86F98D

Proof. We first show that the $m^*$-measurable sets form a $\sigma $-algebra on which $m^*$ is countably additive. The definition is symmetric in $F$ and $X\setminus F$, so complements of measurable sets are measurable. If $F$ and $G$ are measurable and $E\subset X$, splitting $E$ by $F$ and then $E\cap F$ by $G$ gives

where the inequality follows from subadditivity, since $E\setminus (F\cap G)=((E\cap F)\setminus G)\cup (E\setminus F)$. The reverse inequality is again subadditivity, so $F\cap G$ is measurable. If $F_1,F_2,\ldots $ are disjoint measurable sets with union $F$, splitting $E$ successively by $F_1,\ldots ,F_ N$ gives

Letting $N\to \infty $ and using subadditivity, we obtain $m^*(E)\geq m^*(E\cap F)+m^*(E\setminus F)$, so $F$ is measurable, and the choice $E=F$ gives $m^*(F)=\sum _ jm^*(F_ j)$. As every countable union of measurable sets is a disjoint union of measurable sets, this proves the claim.

It remains to show that every nonempty closed set $F\subset X$ is measurable. By subadditivity, we only need $m^*(E)\geq m^*(E\cap F)+m^*(E\setminus F)$, and only for sets $E\subset X$ with $m^*(E)<\infty $. For $N\geq 1$ put $E_ N=\{ x\in E:\operatorname {dist}(x,F)\geq N^{-1}\} $. Since $\operatorname {dist}(E\cap F,E_ N)\geq N^{-1}$, the hypothesis gives $m^*(E)\geq m^*(E\cap F)+m^*(E_ N)$, and it suffices to show that $m^*(E_ N)\to m^*(E\setminus F)$. Consider the layers

Since $\operatorname {dist}(\cdot ,F)$ is $1$-Lipschitz, $\operatorname {dist}(A_ k,A_{k'})\geq (k+1)^{-1}-(k+2)^{-1}>0$ whenever $k'\geq k+2$. Hence finitely many layers whose indices have the same parity are pairwise at positive distance, and repeated use of the hypothesis shows that the sum of their outer measures is at most $m^*(E)$. Consequently $\sum _ km^*(A_ k)\leq 2m^*(E)<\infty $. Since $F$ is closed, $E\setminus F\subset E_ N\cup \bigcup _{k\geq N}A_ k$, and therefore

The tails of the convergent series tend to zero, so $m^*(E_ N)\to m^*(E\setminus F)$. Thus the $\sigma $-algebra of $m^*$-measurable sets contains the closed sets, and hence all Borel sets. $\square$

Proof. We construct $\mu $ from a metric outer measure (lemma 2.3.2) and identify $L$ with integration against $\mu $ by slicing functions into level sets. Call $f$ admissible for an open set $U\subset \Omega $ if $f\in C_ c(\Omega ,\mathbb R)$, $0\leq f\leq 1$ and $\operatorname{Supp}f\subset U$, and put

Then $m(\varnothing )=0$, and $m$ is monotone.

Step 1: $m$ is countably subadditive on open sets and additive on disjoint ones. Let $U_1,U_2,\ldots $ be open and let $f$ be admissible for $U=\bigcup _ jU_ j$. Covering $\operatorname{Supp}f$ by finitely many open balls whose closures lie in some $U_ j$, we find $n$ and compact, possibly empty, sets $K_ j\subset U_ j$, $1\leq j\leq n$, which cover $\operatorname{Supp}f$. If $\eta _ j$ is a cutoff function for $K_ j$ in $U_ j$, then $\eta =\sum _{j=1}^ n\eta _ j\geq 1$ on $\operatorname{Supp}f$, so $\eta >0$ on an open neighborhood $W$ of $\operatorname{Supp}f$. The functions $f_ j$ equal to $f\eta _ j/\eta $ on $W$ and to $0$ on $\Omega \setminus \operatorname{Supp}f$ are continuous, since both formulas give $0$ where these open sets overlap; moreover $0\leq f_ j\leq f$, $\operatorname{Supp}f_ j\subset U_ j$ and $\sum _ jf_ j=f$. Hence $L(f)=\sum _{j=1}^ nL(f_ j)\leq \sum _ jm(U_ j)$, and so $m(U)\leq \sum _ jm(U_ j)$. If the $U_ j$ are pairwise disjoint and $h_ j$ is admissible for $U_ j$, then the $h_ j$ have disjoint supports, so $\sum _{j\leq N}h_ j$ is admissible for $U$; taking suprema shows that $m$ is countably additive on disjoint open sets.

Step 2: construction of $\mu $. For $A\subset \Omega $ put $m^*(A)=\inf \{ m(U):U\supset A\text{ open}\} $. Then $m^*(U)=m(U)$ for open $U$ by monotonicity of $m$. If $\sum _ jm^*(A_ j)<\infty $ (otherwise there is nothing to prove) and $\varepsilon >0$, open sets $U_ j\supset A_ j$ with $m(U_ j)<m^*(A_ j)+\varepsilon 2^{-j}$ give $m^*(\bigcup _ jA_ j)\leq \sum _ jm^*(A_ j)+\varepsilon $ by Step 1. Thus $m^*$ is an outer measure. Let $A,B\subset \Omega $ be nonempty with $\operatorname {dist}(A,B)=2r>0$, and let $W\supset A\cup B$ be open. The sets $W\cap \{ \operatorname {dist}(\cdot ,A)<r\} $ and $W\cap \{ \operatorname {dist}(\cdot ,B)<r\} $ are open neighborhoods of $A$ and $B$ in $W$, which are disjoint by the triangle inequality, so Step 1 gives $m(W)\geq m^*(A)+m^*(B)$. Taking the infimum over $W$, we obtain the hypothesis of lemma 2.3.2 (if $A$ or $B$ is empty, it holds because $m^*(\varnothing )=0$), and that lemma yields a Borel measure $\mu $ with $\mu (U)=m(U)$ for open $U$.

It remains to show that $\mu $ is finite on compact sets. Every compact set $K\subset \Omega $ has a relatively compact open neighborhood $V$ in $\Omega $, for instance the set $V$ used in the construction of cutoff functions in section 2.2. If $\eta $ is a cutoff function for $\overline V$ in $\Omega $, then (2.3.1) gives $L(f)\leq L(\eta )$ for every $f$ admissible for $V$, so $\mu (K)\leq m(V)\leq L(\eta )<\infty $. By proposition 2.2.2, $\mu $ is a Radon measure.

Step 3: $L(f)=\int f\, \mathrm d\mu $. By linearity, it suffices to treat $f\geq 0$ with $M=\| f\| _\infty >0$. Fix an integer $N\geq 1$, put $\delta =M/N$, and let $U_ k=\{ f>k\delta \} $ for $k\geq 0$, so that $\mu (U_ k)\leq \mu (\operatorname{Supp}f)<\infty $. If $h_ k$ is admissible for $U_ k$, $1\leq k\leq N$, then $h_ k\leq \mathbf1_{U_ k}$, and since at most $f(x)/\delta $ indices $k\geq 1$ satisfy $f(x)>k\delta $, we have $\delta \sum _{k=1}^ Nh_ k\leq \delta \sum _{k=1}^ N\mathbf1_{U_ k}\leq f$. Positivity of $L$ and the supremum over the $h_ k$ give the lower bound

For an upper bound, put $g_ k=\min \{ 1,\max \{ 0,f/\delta -(k-1)\} \} $ for $1\leq k\leq N$; these are continuous with values in $[0,1]$, and $f=\delta \sum _{k=1}^ Ng_ k$. If $\eta $ is a cutoff function for $\operatorname{Supp}f$ in $\Omega $, then $g_1\leq \eta $, so $L(g_1)\leq L(\eta )$. For $k\geq 2$, $g_ k$ vanishes outside the compact set $\{ f\geq (k-1)\delta \} \subset U_{k-2}$, so it is admissible for $U_{k-2}$ and $L(g_ k)\leq \mu (U_{k-2})$. Consequently

and the upper bound exceeds the lower bound by

On the other hand, the simple function $s_ N=\delta \sum _{k=1}^ N\mathbf1_{U_ k}$ satisfies $0\leq f-s_ N\leq \delta $ and vanishes outside $\operatorname{Supp}f$, so $\int f\, \mathrm d\mu $ exceeds the lower bound $\int s_ N\, \mathrm d\mu $ by at most $\delta \mu (\operatorname{Supp}f)$. Hence $|L(f)-\int f\, \mathrm d\mu |\leq \delta (L(\eta )+\mu (\operatorname{Supp}f))$, and letting $N\to \infty $ gives $L(f)=\int f\, \mathrm d\mu $.

Step 4: uniqueness. Let $\nu $ be any Radon measure on $\Omega $ and $U\subset \Omega $ open. If $K\subset U$ is compact and $\eta $ is a cutoff function for $K$ in $U$, then $\nu (K)\leq \int \eta \, \mathrm d\nu \leq \nu (U)$, so inner regularity gives

Hence two Radon measures inducing the same functional agree on open sets, and by outer regularity on all Borel sets. $\square$

Proof. Let $u\in L^ p(\mu )$ and let $(K_ j)$ be a compact exhaustion of $\Omega $. The functions $u\mathbf1_{K_ j\cap \{ |u|\leq j\} }$ converge to $u$ in $L^ p(\mu )$ by dominated convergence (proposition 2.2.3), so it suffices to approximate a bounded Borel function $v$ vanishing outside a compact set $K$. A uniform approximation of $\operatorname{Re}v$ and $\operatorname{Im}v$ by simple functions vanishing outside $K$, with error $\delta $, has $L^ p(\mu )$ error at most $2\delta \mu (K)^{1/p}$; we are thus reduced to indicator functions $\mathbf1_ E$ of Borel sets $E\subset K$. Given $\varepsilon >0$, regularity of $\mu $ gives a compact set $K'\subset E$ and an open set $U\supset E$ with $\mu (U\setminus K')<\varepsilon $. A cutoff function $\eta $ for $K'$ in $U$ satisfies $|\eta -\mathbf1_ E|\leq \mathbf1_{U\setminus K'}$, so $\| \eta -\mathbf1_ E\| _{L^ p(\mu )}^ p<\varepsilon $. $\square$