Proof of Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets

Throughout, continuous maps between metric spaces are Borel measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iv)), and Lipschitz maps, constant maps into $\mathbb{R}$, and sums, real multiples and products of continuous real-valued maps are continuous (A Lipschitz Map is Uniformly Continuous and Continuous (ii), Continuity of Sums and Products of Real-Valued Functions on a Metric Space (i), Continuity of Sums and Products of Real-Valued Functions on a Metric Space (ii), Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iv), Continuity of Sums and Products of Real-Valued Functions on a Metric Space (iii)); for a finite measure $\mu$ on $(S,\mathcal{B}(S))$ and a real $M\ge0$, the constant function $M=M\mathbf{1}_{S}$ lies in $\mathcal{L}^{1}(\mu)$ with integral $M\mu(S)$ (Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Basic Properties of the Integral of Nonnegative Measurable Maps (iv), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)), a measurable function dominated by an integrable one is integrable, the integral is monotone, and $|\int f\,d\mu|\le\int|f|\,d\mu$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (v), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ix), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x)); and the distance to a nonempty closed set $F$ is nonnegative, vanishes on $F$, is positive off $F$ and satisfies $|\operatorname{dist}(x,F)-\operatorname{dist}(y,F)|\le d(x,y)$ (The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (i), The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (v), The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (ii)).

Step 1 ((i)). A function in $C_{b}(S)$ is continuous, hence measurable.

Step 2 ((ii)). Since $f$ is measurable and $|f|\le M$, $f$ is dominated by the integrable constant $M$, so $f\in\mathcal{L}^{1}(\mu)$ and

$$\Bigl|\int_{S}f\,d\mu\Bigr|\le\int_{S}|f|\,d\mu\le\int_{S}M\,d\mu=M\mu(S).$$

Step 3 ((iii)). Constants, $f+g$, $af$ and $fg$ are continuous; a constant $c$ is bounded by $|c|$, and if $|f|\le M$ and $|g|\le N$, then $|f+g|\le M+N$, $|af|\le|a|M$ and $|fg|\le MN$.

Step 4 ((iv)). Let $|f|\le M$ with $M\ge0$. The function $\varphi$ is continuous on the closed interval $[-M,M]$, so by the extreme value theorem, Extreme Value Theorem on a Closed Real Interval (i), there is $K$ with $|\varphi(t)|\le K$ for $t\in[-M,M]$; hence $|\varphi(f(x))|\le K$ for every $x\in S$. The composite $\varphi\circ f$ is continuous by Continuity of Identity Maps, Compositions and Reciprocals (ii).

Step 5 ((v)). The composite $f\circ h$ is continuous by Continuity of Identity Maps, Compositions and Reciprocals (ii), and $|f(h(y))|\le M$ for every $y\in T$ when $|f|\le M$.

Step 6 ((vi)). A Lipschitz function is continuous.

Step 7 ((vii)). For real $u,v$ we have $|\max\{0,u\}-\max\{0,v\}|\le|u-v|$ (if $u,v\ge0$ both sides agree, if $u,v\le0$ the left side is $0$, and if $u>0\ge v$ the left side is $u\le u-v$; the remaining case is symmetric). With $u=1-k\operatorname{dist}(x,F)$, $v=1-k\operatorname{dist}(y,F)$ and the Lipschitz estimate for the distance (The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (ii)), for all $x,y\in S$

$$|f_{k}(x)-f_{k}(y)|\le k\,|\operatorname{dist}(x,F)-\operatorname{dist}(y,F)|\le k\,d(x,y),$$

so $f_{k}$ is Lipschitz with constant $k$. Since $\operatorname{dist}(x,F)\ge0$, $0\le f_{k+1}(x)\le f_{k}(x)\le1$. If $x\in F$, then $\operatorname{dist}(x,F)=0$, so $f_{k}(x)=1=\mathbf{1}_{F}(x)$ for every $k$. If $x\notin F$, then $\operatorname{dist}(x,F)>0$, and by the Archimedean property (Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema (ii)) there is $K\in\mathbb{N}$ with $K\operatorname{dist}(x,F)>1$; for every $k\ge K$, $k\operatorname{dist}(x,F)\ge K\operatorname{dist}(x,F)>1$, so $1-k\operatorname{dist}(x,F)<0$ and $f_{k}(x)=0=\mathbf{1}_{F}(x)$. In both cases $(f_{k}(x))_{k}$ is eventually constant equal to $\mathbf{1}_{F}(x)$, hence converges to it by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i) and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (v).