Proof of Riesz-Markov Representation Theorem on a Compact Metric Space

Throughout, $\mathcal{C}$, $\mathbf{1}$, $\mathcal{T}_d$, $\mathcal{B}(K)$ and $\Lambda$ are as in the statement, $\mathbb{N}$ is the set of natural numbers, $[r]$ is the initial segment determined by $r\in\mathbb{N}$, $\iota:\mathbb{N}\to\mathbb{R}$ is the canonical map of $\mathbb{R}$, and sums $\sum_{k=1}^{n}$ are the finite sums of that definition. Closedness of a subset of $K$ always means closedness in $(K,\mathcal{T}_d)$. Write $\mathbf{0}$ for the constant map on $K$ with value $0$, write $2=1+1$, and put $L=\Lambda(\mathbf{1})$.

Step 1 (elementary properties). Taking $c=0$ in hypothesis (ii) gives $\Lambda(\mathbf{0})=\Lambda(0\cdot\mathbf{0})=0\cdot\Lambda(\mathbf{0})=0$.

(1a) (Monotonicity of $\Lambda$) If $f,g\in\mathcal{C}$ and $f(x)\le g(x)$ for every $x\in K$, then $\Lambda(f)\le\Lambda(g)$. Indeed $g+(-1)f$ lies in $\mathcal{C}$ and is nonnegative at every point, so hypotheses (iii), (i) and (ii) give $0\le\Lambda(g+(-1)f)=\Lambda(g)+(-1)\Lambda(f)$.

(1b) $0\le L$, by hypothesis (iii) applied to $\mathbf{1}$.

(1c) If $f\in\mathcal{C}$ and $M$ is a real number with $0\le M$ and $|f(x)|\le M$ for every $x\in K$, then $|\Lambda(f)|\le M\,L$. Indeed claim 6 of Properties of the Absolute Value in an Ordered Field gives $-M\le f(x)\le M$ for every $x$, so comparing $f$ with the constant maps $(-M)\mathbf{1}$ and $M\mathbf{1}$ and using (1a) and (ii) gives $-M L\le\Lambda(f)\le M L$; claim 6 of that lemma again gives the assertion.

(1d) (Monotonicity of finite sums) If $r\in\mathbb{N}$ and $a,b:[r]\to\mathbb{R}$ satisfy $a_k\le b_k$ for every $k\in[r]$, then $\sum_{k=1}^{r}a_k\le\sum_{k=1}^{r}b_k$. Indeed the family $c$ with $c_k=b_k-a_k$ is nonnegative, so $0\le\sum_{k=1}^{r}c_k$ by claim 5 of Properties of Finite Sums, while claim 2 of that lemma gives $\sum_{k=1}^{r}b_k=\sum_{k=1}^{r}a_k+\sum_{k=1}^{r}c_k$.

(1e) (Approximation principle) If $u,v\in\mathbb{R}$ and $u\le v+\varepsilon$ for every real $\varepsilon>0$, then $u\le v$. Indeed, if $v<u$ then $\varepsilon=(u-v)\cdot 2^{-1}$ is positive and satisfies $\varepsilon<u-v$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, so $v+\varepsilon<u$, contradicting the hypothesis.

Step 2 (admissible functions, and a functional on open sets). Let $U\in\mathcal{T}_d$. We say that a map $f$ is admissible for $U$ if $f\in\mathcal{C}$, $0\le f(x)\le 1$ for every $x\in K$, and there is a closed subset $D\subseteq U$ of $K$ such that $f(x)=0$ for every $x\in K$ with $x\notin D$; such a set $D$ is called a witness for $f$.

Let $\Sigma(U)=\{t\in\mathbb{R}: t=\Lambda(f)\text{ for some }f\text{ admissible for }U\}$. The map $\mathbf{0}$ is admissible for $U$ with witness $\varnothing$, which is closed because $K\setminus\varnothing=K$ lies in $\mathcal{T}_d$; hence $0\in\Sigma(U)$ and $\Sigma(U)$ is nonempty. Every admissible $f$ satisfies $f(x)\le 1$ for all $x$, so $\Lambda(f)\le L$ by (1a); hence $\Sigma(U)$ is bounded above and

exists in $\mathbb{R}$ and satisfies $0\le\lambda(U)\le L$.

(2a) (Monotonicity) If $U,V\in\mathcal{T}_d$ and $U\subseteq V$, then every $f$ admissible for $U$ is admissible for $V$ with the same witness, so $\Sigma(U)\subseteq\Sigma(V)$ and $\lambda(U)\le\lambda(V)$.

(2b) $\lambda(\varnothing)=0$: a witness $D\subseteq\varnothing$ must be $\varnothing$, so the only map admissible for $\varnothing$ is $\mathbf{0}$ and $\Sigma(\varnothing)=\{0\}$.

(2c) $\lambda(K)=L$: the map $\mathbf{1}$ is admissible for $K$ with witness $K$, which is closed because $K\setminus K=\varnothing\in\mathcal{T}_d$, so $L\in\Sigma(K)$.

Step 3 (finite subadditivity on open sets). Let $n\in\mathbb{N}$, let $U_i\in\mathcal{T}_d$ for $i\in[n]$, and put $U=\bigcup_{i\in[n]}U_i$, which lies in $\mathcal{T}_d$. We claim that

Let $f$ be admissible for $U$ with witness $D$. Then $D$ is a closed subset of the compact space $K$, so $D$ is compact in $(K,\mathcal{T}_d)$ by Closed Subset of a Compact Space is Compact, and $D\subseteq\bigcup_{i\in[n]}U_i$. By Continuous Partition of Unity Subordinate to a Finite Open Cover of a Compact Set in a Metric Space, applied with the metric space $(K,d)$, the compact set $D$ and the family $(U_i)_{i\in[n]}$, there are maps $h_i:K\to\mathbb{R}$ and closed subsets $D_i\subseteq U_i$, one for each $i\in[n]$, such that each $h_i$ is continuous on $K$ with $0\le h_i(x)\le 1$, each $h_i$ vanishes at every point outside $D_i$, $\sum_{i=1}^{n}h_i(x)\le 1$ for every $x\in K$, and $\sum_{i=1}^{n}h_i(x)=1$ for every $x\in D$.

For $i\in[n]$ let $h_if$ denote the pointwise product, which lies in $\mathcal{C}$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. It satisfies $0\le h_i(x)f(x)\le 1$ for every $x$, and $h_i(x)f(x)=0$ whenever $x\notin D_i$; hence $h_if$ is admissible for $U_i$ with witness $D_i$, and $\Lambda(h_if)\le\lambda(U_i)$.

Moreover $f(x)=\sum_{i=1}^{n}h_i(x)f(x)$ for every $x\in K$. Indeed claim 3 of Properties of Finite Sums gives $\sum_{i=1}^{n}f(x)h_i(x)=f(x)\sum_{i=1}^{n}h_i(x)$, which equals $f(x)$ when $x\in D$ and equals $0=f(x)$ when $x\notin D$.

By hypothesis (i), claim 1 of Properties of Finite Sums and induction on $n$ we get $\Lambda(f)=\sum_{i=1}^{n}\Lambda(h_if)$, and (1d) then gives $\Lambda(f)\le\sum_{i=1}^{n}\lambda(U_i)$. As $f$ was an arbitrary map admissible for $U$, the right-hand side is an upper bound for $\Sigma(U)$, so $\lambda(U)\le\sum_{i=1}^{n}\lambda(U_i)$.

Step 4 (countable subadditivity on open sets). Let $(U_m)_{m\in\mathbb{N}}$ be a family of subsets of $K$ with $U_m\in\mathcal{T}_d$ for every $m$, and put $U=\bigcup_{m\in\mathbb{N}}U_m\in\mathcal{T}_d$. We claim that

the right-hand side being the sum of a sequence in $[0,\infty]$ as fixed in that definition.

If that sum is $\infty$ the inequality holds because $\lambda(U)$ is real and every nonnegative real is smaller than $\infty$. So assume it is real; it is then the least upper bound of the partial sums $\sum_{m=1}^{n}\lambda(U_m)$, all of which are nonnegative by claim 5 of Properties of Finite Sums.

Let $f$ be admissible for $U$ with witness $D$, which is compact as in Step 3. Since $D\subseteq\bigcup_{m\in\mathbb{N}}U_m$ and every $U_m$ lies in $\mathcal{T}_d$, compactness of $D$ gives a finite subset $J\subseteq\mathbb{N}$ with $D\subseteq\bigcup_{m\in J}U_m$.

If $J$ is empty then $D$ is empty, so $f=\mathbf{0}$ and $\Lambda(f)=0\le\sum_{m\in\mathbb{N}}\lambda(U_m)$. Otherwise $J$ has $p$ elements for some $p\in\mathbb{N}$, and a bijection $[p]\to J$ is a $p$-tuple $c$ in $\mathbb{N}$ whose set of components is $J$. The order on $\mathbb{N}$ is a total order, by claims 1, 2 and 3 of Properties of the Order on the Natural Numbers, so Greatest Element of a Finite Family in a Totally Ordered Set gives $j\in[p]$ with $c_k\le c_j$ for every $k\in[p]$. Put $n=c_j$. Every $m\in J$ satisfies $m\le n$, hence $m\in[n]$, so $D\subseteq\bigcup_{m\in[n]}U_m$ and $f$ is admissible for that union. Step 3 now gives

the last inequality because a partial sum is at most the least upper bound of the partial sums. Taking the supremum over admissible $f$ gives the claim.

Step 5 (an outer measure). For $A\subseteq K$ let $\Theta(A)=\{t\in\mathbb{R}: t=\lambda(V)\text{ for some }V\in\mathcal{T}_d\text{ with }A\subseteq V\}$. This set is nonempty, since $K\in\mathcal{T}_d$, and $0$ is a lower bound for it, so

exists by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below, and $0\le\mu^{*}(A)\le\lambda(K)=L$.

(5a) $\mu^{*}(\varnothing)=0$, because $\varnothing\in\mathcal{T}_d$ and $\lambda(\varnothing)=0$ by (2b).

(5b) (Monotonicity) If $A\subseteq B\subseteq K$ then $\Theta(B)\subseteq\Theta(A)$, so every lower bound for $\Theta(A)$ is a lower bound for $\Theta(B)$ and $\mu^{*}(A)\le\mu^{*}(B)$.

(5c) $\mu^{*}(V)=\lambda(V)$ for every $V\in\mathcal{T}_d$: on one hand $\lambda(V)\in\Theta(V)$, so $\mu^{*}(V)\le\lambda(V)$; on the other hand $\lambda(V)$ is a lower bound for $\Theta(V)$ by (2a), so $\lambda(V)\le\mu^{*}(V)$.

(5d) (Countable subadditivity) Let $(A_m)_{m\in\mathbb{N}}$ be a family of subsets of $K$ and $A=\bigcup_{m\in\mathbb{N}}A_m$. If $\sum_{m\in\mathbb{N}}\mu^{*}(A_m)=\infty$ there is nothing to prove, so assume it is real.

Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Define a sequence $(\varepsilon_m)_{m\in\mathbb{N}}$ in $\mathbb{R}$ recursively by $\varepsilon_1=\varepsilon\cdot 2^{-1}$ and $\varepsilon_{m+1}=\varepsilon_m\cdot 2^{-1}$. By claim 8 of Elementary Order Arithmetic in an Ordered Field and induction, $0<\varepsilon_m$ and $\varepsilon_{m+1}+\varepsilon_{m+1}=\varepsilon_m$ for every $m$, and $\varepsilon_1+\varepsilon_1=\varepsilon$. A further induction on $n$, using claim 1 of Properties of Finite Sums, gives

and since $0<\varepsilon_n$, claim 1 of Elementary Order Arithmetic in an Ordered Field gives $\varepsilon-\varepsilon_n<\varepsilon$.

For each $m\in\mathbb{N}$ let $\mathcal{V}_m=\{V\in\mathcal{T}_d: A_m\subseteq V\text{ and }\lambda(V)<\mu^{*}(A_m)+\varepsilon_m\}$. By claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to the set $\Theta(A_m)$ and the positive number $\varepsilon_m$, there is an element of $\Theta(A_m)$ smaller than $\mu^{*}(A_m)+\varepsilon_m$, and by the description of $\Theta(A_m)$ that element is $\lambda(V)$ for some $V\in\mathcal{T}_d$ containing $A_m$; hence $\mathcal{V}_m$ is nonempty. By Axiom of Countable Choice, applied to the family $(\mathcal{V}_m)_{m\in\mathbb{N}}$ of subsets of $\mathcal{T}_d$, there is a sequence $(V_m)_{m\in\mathbb{N}}$ in $\mathcal{T}_d$ with $V_m\in\mathcal{V}_m$ for every $m$.

Then $A\subseteq\bigcup_{m\in\mathbb{N}}V_m$, so (5b), (5c) and Step 4 give $\mu^{*}(A)\le\lambda\bigl(\bigcup_{m\in\mathbb{N}}V_m\bigr)\le\sum_{m\in\mathbb{N}}\lambda(V_m)$. For every $n\in\mathbb{N}$, (1d) and claim 2 of Properties of Finite Sums give

In particular the partial sums of $(\lambda(V_m))_{m\in\mathbb{N}}$ are bounded above, so $\sum_{m\in\mathbb{N}}\lambda(V_m)$ is real and is their least upper bound; hence it is at most $\sum_{m\in\mathbb{N}}\mu^{*}(A_m)+\varepsilon$, and therefore $\mu^{*}(A)\le\sum_{m\in\mathbb{N}}\mu^{*}(A_m)+\varepsilon$. As $\varepsilon>0$ was arbitrary, (1e) gives $\mu^{*}(A)\le\sum_{m\in\mathbb{N}}\mu^{*}(A_m)$.

By (5a), (5b) and (5d), $\mu^{*}$ is an outer measure on $K$.

Step 6 (open sets are Caratheodory measurable). Let $U\in\mathcal{T}_d$ and let $A\subseteq K$. Apply (5d) to the family $(A_m)_{m\in\mathbb{N}}$ with $A_1=A\cap U$, $A_2=A\setminus U$ and $A_m=\varnothing$ for every $m$ with $2<m$, whose union is $A$. By (5a) every term of the sequence $(\mu^{*}(A_m))_{m\in\mathbb{N}}$ beyond the second is $0$, so by claim 1 of Properties of Finite Sums an induction on $n$ shows that the partial sum $\sum_{m=1}^{n}\mu^{*}(A_m)$ equals $\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)$ for every $n$ with $2\le n$. These partial sums are therefore bounded above, and their least upper bound is that same constant, so by the convention fixed for the sum of a sequence in $[0,\infty]$ we get $\sum_{m\in\mathbb{N}}\mu^{*}(A_m)=\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)$ and hence $\mu^{*}(A)\le\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)$.

For the reverse inequality, let $V\in\mathcal{T}_d$ with $A\subseteq V$, and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. The set $V\cap U$ lies in $\mathcal{T}_d$, so by claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is a map $f$ admissible for $V\cap U$ with $\lambda(V\cap U)-\varepsilon<\Lambda(f)$; let $D$ be a witness for $f$, so $D$ is closed, $D\subseteq V\cap U$, and $f$ vanishes outside $D$.

The set $V\setminus D$ is the intersection of $V$ with $K\setminus D$, both of which lie in $\mathcal{T}_d$, so $V\setminus D\in\mathcal{T}_d$ by the definition of a topological space. Again by claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is a map $g$ admissible for $V\setminus D$ with $\lambda(V\setminus D)-\varepsilon<\Lambda(g)$; let $E$ be a witness for $g$, so $E$ is closed, $E\subseteq V\setminus D$, and $g$ vanishes outside $E$.

The map $f+g$ lies in $\mathcal{C}$ and is nonnegative at every point. Since $E\subseteq V\setminus D$ we have $D\cap E=\varnothing$, so for every $x\in K$ at least one of $f(x)$ and $g(x)$ is $0$: if $x\notin D$ then $f(x)=0$, while if $x\in D$ then $x\notin E$ and $g(x)=0$. Hence $f(x)+g(x)\le 1$ for every $x$. The set $D\cup E$ is closed, because $K\setminus(D\cup E)=(K\setminus D)\cap(K\setminus E)$ is an intersection of two members of $\mathcal{T}_d$; it is contained in $V$; and $f+g$ vanishes outside it. So $f+g$ is admissible for $V$, and by hypothesis (i),

Since $D\subseteq U$ we have $V\setminus U\subseteq V\setminus D$, so (5c) and (5b) give $\lambda(V\setminus D)=\mu^{*}(V\setminus D)\ge\mu^{*}(V\setminus U)\ge\mu^{*}(A\setminus U)$, and likewise $\lambda(V\cap U)=\mu^{*}(V\cap U)\ge\mu^{*}(A\cap U)$. Writing $q=\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)$ we obtain $q-\varepsilon-\varepsilon<\lambda(V)$ for every real $\varepsilon>0$. If $\lambda(V)<q$ then, putting $\varepsilon=(q-\lambda(V))\cdot 2^{-1}\cdot 2^{-1}$, claim 8 of Elementary Order Arithmetic in an Ordered Field gives $0<\varepsilon$ and $\varepsilon+\varepsilon<q-\lambda(V)$, hence $\lambda(V)<q-\varepsilon-\varepsilon$, a contradiction. Therefore $q\le\lambda(V)$.

Since $V$ was an arbitrary member of $\mathcal{T}_d$ containing $A$, the number $q$ is a lower bound for $\Theta(A)$, so $q\le\mu^{*}(A)$. Combined with the first inequality this gives $\mu^{*}(A)=\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)$, so $U$ is Caratheodory measurable with respect to $\mu^{*}$ in the sense of that definition.

Step 7 (the Borel measure). Let $\mathcal{M}$ be the family of all subsets of $K$ that are Caratheodory measurable with respect to $\mu^{*}$. By Caratheodory Extension Theorem, $\mathcal{M}$ is a $\sigma$-algebra on $K$ and the restriction of $\mu^{*}$ to $\mathcal{M}$ is a measure on $(K,\mathcal{M})$. By Step 6, $\mathcal{T}_d\subseteq\mathcal{M}$, so by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra the $\sigma$-algebra generated by $\mathcal{T}_d$, which is $\mathcal{B}(K)$, satisfies $\mathcal{B}(K)\subseteq\mathcal{M}$.

Let $\mu$ be the restriction of $\mu^{*}$ to $\mathcal{B}(K)$. Then $\mu(\varnothing)=0$, and for every sequence of pairwise disjoint members of $\mathcal{B}(K)$ the countable additivity required of a measure holds because those sets and their union lie in $\mathcal{M}$, where the restriction of $\mu^{*}$ is a measure. Hence $\mu$ is a measure on $(K,\mathcal{B}(K))$, that is, a Borel measure on $(K,d)$. By (5c) and (2c), $\mu(K)=\lambda(K)=L$, which is a real number, so $\mu$ is finite.

Two consequences are used below. First, $\mu(V)=\lambda(V)$ for every $V\in\mathcal{T}_d$, by (5c). Second, for every $B\in\mathcal{B}(K)$ and every real $\eta>0$ there is $V\in\mathcal{T}_d$ with $B\subseteq V$ and $\mu(V)<\mu(B)+\eta$; this is claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ applied to $\Theta(B)$, together with (5c).

Step 8 (integrability of continuous maps). Let $f\in\mathcal{C}$. Since $K$ is nonempty and compact, Extreme Value Theorem on a Compact Subset of a Metric Space gives $x_{\min},x_{\max}\in K$ with $f(x_{\min})\le f(x)\le f(x_{\max})$ for every $x\in K$. Put $M_f=\max\{|f(x_{\min})|,|f(x_{\max})|\}$, the maximum of two elements, so $0\le M_f$ by claim 1 of Properties of the Absolute Value in an Ordered Field, and by claims 3 and 6 of that lemma $|f(x)|\le M_f$ for every $x\in K$. By claims 2 and 3 of Borel Measurability and Bounded Integration on a Metric Space, $f$ is measurable with respect to $\mathcal{B}(K)$ and the Borel $\sigma$-algebra of the real line, so by part (b) of claim 6 of that lemma $f$ is integrable with respect to $\mu$.

Step 9 (the representation, reduced to one inequality). We claim that

Granting $(\ast)$, apply it to $(-1)f\in\mathcal{C}$: hypothesis (ii) and claim 2 of Linearity and Monotonicity of the Lebesgue Integral give $-\Lambda(f)=\Lambda((-1)f)\le\int_K(-1)f\,d\mu=-\int_Kf\,d\mu$, hence $\int_Kf\,d\mu\le\Lambda(f)$. Together with $(\ast)$ this proves claim 1 of the theorem, the remaining assertions of that claim having been established in Steps 7 and 8.

We may assume $f$ nonnegative. Indeed, let $f\in\mathcal{C}$ with $M_f$ as in Step 8 and put $F=f+M_f\mathbf{1}$, so that $F\in\mathcal{C}$ and $0\le F(x)$ for every $x\in K$. Hypotheses (i) and (ii) give $\Lambda(F)=\Lambda(f)+M_fL$, while part (a) of claim 6 of Borel Measurability and Bounded Integration on a Metric Space and claim 2 of Linearity and Monotonicity of the Lebesgue Integral give $\int_KF\,d\mu=\int_Kf\,d\mu+M_f\,\mu(K)=\int_Kf\,d\mu+M_fL$. Hence $(\ast)$ holds for $f$ if and only if it holds for $F$.

Step 10 (proof of $(\ast)$ for nonnegative $f$). Let $f\in\mathcal{C}$ with $0\le f(x)$ for every $x\in K$, let $M_f$ be as in Step 8 and put $b=M_f+1$, so that $0\le f(x)<b$ for every $x\in K$. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$.

Put $\delta=\varepsilon\cdot 2^{-1}\cdot 2^{-1}$; by claim 8 of Elementary Order Arithmetic in an Ordered Field, $0<\delta$ and $\delta+\delta=\varepsilon\cdot 2^{-1}<\varepsilon$. By claim 2 of The Archimedean Property of the Real Numbers choose $r\in\mathbb{N}$ with $b<\iota(r)\,\delta$. Put $y_0=-\delta$ and $y_i=\iota(i)\,\delta$ for $i\in[r]$; here and below, for $i\in[r]$ with $i\ne 1$ we write $i-1$ for the unique natural number whose successor is $i$, which exists by claim 6 of Arithmetic of Addition on the Natural Numbers. By claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $0<\iota(i)$, so $0<y_i$ by claim 5 of Elementary Order Arithmetic in an Ordered Field; by claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $y_i=y_{i-1}+\delta$ for $i\in[r]$ with $i\ne 1$; and by claim 6 of that lemma together with claim 10 of Elementary Order Arithmetic in an Ordered Field we have $y_i\le y_r$ for every $i\in[r]$.

For $i\in[r]$ put $E_i=\{x\in K: y_{i-1}<f(x)\le y_i\}$. We record four facts.

(10a) $y_i-\varepsilon<f(x)$ for every $x\in E_i$. For $i=1$ this holds because $y_1-\varepsilon=\delta-\varepsilon<-\delta\le 0\le f(x)$, using $\delta+\delta<\varepsilon$. For $i\ne 1$ it holds because $y_i-\varepsilon<y_i-\delta-\delta<y_i-\delta=y_{i-1}<f(x)$.

(10b) The sets $E_i$ are pairwise disjoint and $\bigcup_{i\in[r]}E_i=K$. Disjointness: if $i,j\in[r]$ and $i<j$ then $i\le j-1$, so $y_i\le y_{j-1}$, and a point of $E_i$ satisfies $f(x)\le y_i\le y_{j-1}$ while a point of $E_j$ satisfies $y_{j-1}<f(x)$. Covering: given $x\in K$, the set of those $i\in[r]$ with $f(x)\le y_i$ is nonempty, because $f(x)<b<\iota(r)\delta=y_r$; by The Natural Numbers Are Well Ordered it has a least element $i_0$, and then $f(x)\le y_{i_0}$ while $y_{i_0-1}<f(x)$ (for $i_0\ne 1$ by minimality, and for $i_0=1$ because $y_0=-\delta<0\le f(x)$), so $x\in E_{i_0}$.

(10c) $E_i\in\mathcal{B}(K)$ for every $i\in[r]$. Being continuous on $K$, $f$ is both upper and lower semicontinuous on $K$ by claim 2 of Semicontinuity Under Negation and Characterization of Continuity, so by claims 2 and 3 of Semicontinuity via Sublevel and Superlevel Sets, applied with the subset $K$ of $K$ itself, the set $\{x\in K: y_{i-1}<f(x)\}$ lies in $\mathcal{T}_d$ and the set $\{x\in K: f(x)\le y_i\}$ is closed. Both are Borel by claim 1 of Borel Measurability and Bounded Integration on a Metric Space, and $E_i$ is their intersection, which lies in $\mathcal{B}(K)$ because a $\sigma$-algebra is closed under finite intersections.

(10d) $\sum_{i=1}^{r}\mu(E_i)=\mu(K)=L$, by claim 1 of Basic Properties of a Measure together with (10b).

Put $\theta=\varepsilon\cdot\bigl(\iota(r)\,(y_r+\varepsilon)\bigr)^{-1}$, which is positive by claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field, since $0<\iota(r)$ by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field and $0<y_r+\varepsilon$.

For each $i\in[r]$ the set $W_i=\{x\in K: f(x)<y_i+\varepsilon\}$ lies in $\mathcal{T}_d$ by claim 1 of Semicontinuity via Sublevel and Superlevel Sets, and $E_i\subseteq W_i$ because $f(x)\le y_i<y_i+\varepsilon$ on $E_i$. By Step 7 there is $V_i\in\mathcal{T}_d$ with $E_i\subseteq V_i$ and $\mu(V_i)<\mu(E_i)+\theta$; put $U_i=V_i\cap W_i$, so that $U_i\in\mathcal{T}_d$, $E_i\subseteq U_i$, $\mu(U_i)\le\mu(V_i)<\mu(E_i)+\theta$ by claim 2 of Basic Properties of a Measure, and $f(x)<y_i+\varepsilon$ for every $x\in U_i$. Only finitely many selections are made here, so a family $(U_i)_{i\in[r]}$ with these properties exists by induction on $r$ and no choice principle is needed.

By (10b), $K=\bigcup_{i\in[r]}E_i\subseteq\bigcup_{i\in[r]}U_i$. Applying Continuous Partition of Unity Subordinate to a Finite Open Cover of a Compact Set in a Metric Space with the metric space $(K,d)$, the compact set $K$ and the family $(U_i)_{i\in[r]}$ gives maps $h_i:K\to\mathbb{R}$ and closed sets $D_i\subseteq U_i$ with $h_i$ continuous on $K$, $0\le h_i(x)\le 1$, $h_i$ vanishing outside $D_i$, and $\sum_{i=1}^{r}h_i(x)=1$ for every $x\in K$.

As in Step 3, each $h_i$ is admissible for $U_i$ with witness $D_i$, so $\Lambda(h_i)\le\lambda(U_i)=\mu(U_i)$, and $0\le\Lambda(h_i)$ by hypothesis (iii). Also, by claim 3 of Properties of Finite Sums, $\sum_{i=1}^{r}h_i(x)f(x)=f(x)\sum_{i=1}^{r}h_i(x)=f(x)$ for every $x\in K$, so hypothesis (i), claim 1 of Properties of Finite Sums and induction on $r$ give $\Lambda(f)=\sum_{i=1}^{r}\Lambda(h_if)$.

For each $i\in[r]$ and every $x\in K$ we have $h_i(x)f(x)\le h_i(x)(y_i+\varepsilon)$: if $h_i(x)=0$ both sides are $0$, and otherwise $0<h_i(x)$, so $x\in D_i\subseteq U_i$, whence $f(x)<y_i+\varepsilon$, and multiplying by the positive number $h_i(x)$ preserves the inequality by claim 10 of Elementary Order Arithmetic in an Ordered Field. Hence by (1a) and hypothesis (ii), and then by $0<y_i+\varepsilon$,

Summing over $i\in[r]$ and using (1d) and claims 2 and 3 of Properties of Finite Sums,

We bound the two terms. Since $y_i+\varepsilon\le y_r+\varepsilon$ for every $i\in[r]$, (1d) and the identity $\sum_{i=1}^{r}c=\iota(r)\,c$ for a constant family, which follows by induction on $r$ from claim 1 of Properties of Finite Sums and claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, give

For the other term, $y_i+\varepsilon=(y_i-\varepsilon)+(\varepsilon+\varepsilon)$, so claims 2 and 3 of Properties of Finite Sums together with (10d) give

Finally we compare the remaining sum with the integral. For $i\in[r]$ put $t_i=\max\{0,y_i-\varepsilon\}$, the maximum of two elements, and let $u_i:K\to\mathbb{R}$ be the map with $u_i(x)=t_i$ for $x\in E_i$ and $u_i(x)=0$ for $x\in K$ with $x\notin E_i$. Each $u_i$ is nonnegative, is measurable by (10c), and takes at most the two distinct values $t_i$ and $0$, so it is a simple function whose integral, computed from its standard representation as in that definition, is $t_i\,\mu(E_i)+0\cdot\mu(K\setminus E_i)=t_i\,\mu(E_i)$; when $t_i=0$ the map $u_i$ is identically $0$ and its integral is $0=t_i\mu(E_i)$ as well. Let $s=\sum_{i=1}^{r}u_i$, formed pointwise. By (10b) the sets $E_i$ are pairwise disjoint with union $K$, so $s(x)=t_i$ for $x\in E_i$; and by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, claim 1 of Properties of Finite Sums and induction on $r$,

the integrals being those of nonnegative measurable functions in the sense of Lebesgue Integral of a Nonnegative Measurable Function, which agree with the integrals of nonnegative simple functions as that definition records. Moreover $s(x)\le f(x)$ for every $x\in K$: if $x\in E_i$ then $y_i-\varepsilon<f(x)$ by (10a) and $0\le f(x)$, so $t_i\le f(x)$. Hence, using (1d) for the first inequality and claim 1 of Linearity and Monotonicity of the Lebesgue Integral together with part (c) of claim 6 of Borel Measurability and Bounded Integration on a Metric Space for the last,