Proof of The Lebesgue Integral over ℝ of a Nonnegative Continuous Function as the Limit of Its Riemann Integrals over [−n, n]

Throughout, closed intervals are Borel sets (background of the settings), $h|_{[a,b]}$ is continuous on $[a,b]$ when $h$ is continuous on $\mathbb{R}$ (The Real Line: Standing Notation and Background for Calculus on Intervals (vi)), and the functions $h\mathbf{1}_{[-n,n]}$ form a sequence of functions $\mathbb{R}\to\mathbb{R}$ (Intervals of the Real Line as Defined Set Symbols, Typed Terms and Values of Maps (iii)).

Step 1 (proof of (i)). Let $\varphi=g|_{[a,b]}$. By Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii), $\varphi$ is measurable for $\mathcal{B}_{[a,b]}$, $\varphi\in\mathcal{L}^{1}(\lambda_{[a,b]})$ and $\int\varphi\,d\lambda_{[a,b]}=\int_{a}^{b}\varphi$; here $([a,b],\mathcal{B}_{[a,b]},\lambda_{[a,b]})$ is the restriction of $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ to $[a,b]$ of Integrals of Functions over a Measurable Subset (i). The function $g\mathbf{1}_{[a,b]}$ equals $\varphi$ on $[a,b]$ and $0$ off $[a,b]$, so by Integrals over Measurable Subsets (iv) it is measurable, lies in $\mathcal{L}^{1}(\lambda)$, and

$$\int_{\mathbb{R}}g\mathbf{1}_{[a,b]}\,d\lambda=\int\varphi\,d\lambda_{[a,b]}=\int_{a}^{b}g|_{[a,b]} .$$

Step 2 (proof of (ii)). Let $f_{n}=h\mathbf{1}_{[-n,n]}$ for $n\in\mathbb{N}$. Since $0<n$ in $\mathbb{R}$ (Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vi)), $-n<n$, so Step 1 applies with $g=h$, $a=-n$ and $b=n$: $f_{n}$ is measurable and $\int_{\mathbb{R}}f_{n}\,d\lambda=I_{n}$. As $f_{n}$ is nonnegative, Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi) puts it in $\mathcal{M}^{+}(\mathbb{R},\mathcal{B}(\mathbb{R}))$ with

$$\int_{\mathbb{R}}f_{n}\,d\lambda=I_{n}\in[0,\infty)$$

for its integral as an element of $\mathcal{M}^{+}$; so the sequence $(I_{n})$ whose supremum (ii) takes is the sequence $n\mapsto\int_{\mathbb{R}}f_{n}\,d\lambda$ of these integrals. Since $h\ge0$ and $[-n,n]\subseteq[-n-1,n+1]$, $f_{n}(x)\le f_{n+1}(x)\le h(x)$ for every $x\in\mathbb{R}$. Given $x\in\mathbb{R}$, 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)) gives $N_{1},N_{2}\in\mathbb{N}$ with $x<N_{1}$ and $-x<N_{2}$; for $N=N_{1}+N_{2}$ we have $-N<x<N$, as $N_{1},N_{2}>0$, so $x\in[-N,N]$ and $f_{N}(x)=h(x)$. Hence $h(x)=\sup_{n}f_{n}(x)$ for every $x\in\mathbb{R}$. By the monotone convergence theorem, $h\in\mathcal{M}^{+}(\mathbb{R},\mathcal{B}(\mathbb{R}))$ (Monotone Convergence Theorem (i)) and, by Monotone Convergence Theorem (ii),

$$\int_{\mathbb{R}}h\,d\lambda=\sup_{n\in\mathbb{N}}\int_{\mathbb{R}}f_{n}\,d\lambda=\sup_{n\in\mathbb{N}}I_{n} .$$

Step 3 (proof of (iii)). The function $h$ is real-valued, nonnegative and lies in $\mathcal{M}^{+}$ by Step 2, so it is measurable as a map into $\mathbb{R}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)). By Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi), $h\in\mathcal{L}^{1}(\lambda)$ if and only if $\int_{\mathbb{R}}h\,d\lambda<\infty$ for the integral of $h\in\mathcal{M}^{+}$, and then the two integrals of $h$ agree; by Step 2 this is $\sup_{n}I_{n}<\infty$.

Step 4 (proof of (iv)). Let $h\in\mathcal{L}^{1}(\lambda)$. By Step 3, $\int_{\mathbb{R}}h\,d\lambda$, the integral of $h\in\mathcal{M}^{+}$, is a real number and equals the integral of $h\in\mathcal{L}^{1}(\lambda)$. The sequence $(I_{n})$ is nondecreasing, since $I_{n}=\int f_{n}\,d\lambda\le\int f_{n+1}\,d\lambda=I_{n+1}$ by Basic Properties of the Integral of Nonnegative Measurable Maps (iii), and bounded above by $\int_{\mathbb{R}}h\,d\lambda$, since $f_{n}\le h$ gives $I_{n}\le\int_{\mathbb{R}}h\,d\lambda$ by the same monotonicity. By Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences (iii), $I_{n}$ converges to the supremum of $\{I_{n}:n\in\mathbb{N}\}$ in $\mathbb{R}$, which is $\sup_{n}I_{n}$ by Order and Suprema in the Extended Real Line (iv), and so $\int_{\mathbb{R}}h\,d\lambda$ by Step 2.