Theorem 2.3.3 (Riesz representation)0C614771FB98

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. $\square$