Proof of Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval

Claim 1 is exactly Continuous Functions on a Closed Interval are Riemann Integrable. Write $I=\int_a^bh(x)\,dx$ for the Riemann integral of Riemann Integrability on a Closed Interval: for every $\varepsilon>0$ there is $\delta>0$ such that every Riemann sum of $h$ over a tagged partition of mesh less than $\delta$ differs from $I$ by less than $\varepsilon$.

Step 1 (measurability and integrability). Fix $t\in\mathbb{R}$. By continuity, for each $x\in[a,b]$ with $h(x)>t$ there is $\delta_x>0$ with $h>t$ on $(x-\delta_x,x+\delta_x)\cap[a,b]$; with $G=\bigcup_x(x-\delta_x,x+\delta_x)$ (union over such $x$), an open set, $\{x\in[a,b]:h(x)>t\}=G\cap[a,b]$. Hence $\{x\in\mathbb{R}:\tilde{h}(x)>t\}$ equals $G\cap[a,b]$ if $t\ge0$ and $(G\cap[a,b])\cup(\mathbb{R}\setminus[a,b])$ if $t<0$; in both cases a Borel set. The rays $(t,\infty)$ generate $\mathcal{B}(\mathbb{R})$: every open interval with rational endpoints is obtained from rays by countable set operations (e.g. $(c,d)=(c,\infty)\setminus\bigcap_{k\ge1}(d-1/k,\infty)$), and every open set is a countable union of such intervals by density of the rationals; so $\tilde{h}$ is Borel measurable by the generator criterion there, and so is $|\tilde{h}|=\tilde{h}^{+}+\tilde{h}^{-}$ (positive and negative parts as in Integrable Function and the Lebesgue Integral). By Extreme Value Theorem on a Compact Interval, $h$ attains a maximum and a minimum on $[a,b]$, so $M=\sup_{[a,b]}|h|$ is a finite real number and $|\tilde{h}|\le M\mathbf{1}_{[a,b]}$ pointwise. Since the simple function $M\mathbf{1}_{[a,b]}$ has integral $M\,\lambda([a,b])=M(b-a)$ — Lebesgue measure assigns each interval its length — monotonicity (Linearity and Monotonicity of the Lebesgue Integral) gives $\int_{\mathbb{R}}|\tilde{h}|\,d\lambda\le M(b-a)<\infty$, so $\tilde{h}$ is integrable. This proves Claim 2.

Step 2 (reduction to a nonnegative integrand). Let $k=h+M$. Then $k\ge0$ on $[a,b]$ (since $h\ge-|h|\ge-M$ there), and $k$ is continuous on $[a,b]$ (the sum of $h$ and a constant, by limit arithmetic in Continuity on a Closed Interval). On the Riemann side we work directly with Riemann Integrability on a Closed Interval: every Riemann sum of $k$ over a tagged partition equals the corresponding Riemann sum of $h$ plus $M(b-a)$, because $\sum_i(h(\xi_i)+M)(t_i-t_{i-1})=\sum_ih(\xi_i)(t_i-t_{i-1})+M(b-a)$; hence (with the same $\delta$ for each $\varepsilon$) $k$ is Riemann integrable with $\int_a^bk(x)\,dx=I+M(b-a)$. On the Lebesgue side, the zero extension of $k$ is $\tilde{h}+M\mathbf{1}_{[a,b]}$, whose integral is $\int\tilde{h}\,d\lambda+M(b-a)$ by linearity for integrable functions (Linearity and Monotonicity of the Lebesgue Integral). So Claim 3 for $k$ implies Claim 3 for $h$, and we may assume $h\ge0$.

Step 3 (sandwich, valid for every partition). Let $\mathcal{P}:a=t_0<t_1<\dots<t_m=b$ be a partition, and let $m_i$ and $M_i$ be the infimum and supremum of $h$ on $[t_{i-1},t_i]$, so that the lower and upper sums are $L(h,\mathcal{P})=\sum_i m_i(t_i-t_{i-1})$ and $U(h,\mathcal{P})=\sum_i M_i(t_i-t_{i-1})$. The nonnegative simple functions $\varphi=\sum_i m_i\mathbf{1}_{(t_{i-1},t_i]}$ and $\psi=\sum_i M_i\mathbf{1}_{(t_{i-1},t_i]}$ satisfy $\int\varphi\,d\lambda=L(h,\mathcal{P})$ and $\int\psi\,d\lambda=U(h,\mathcal{P})$: the displayed representations need not be the standard representations of Simple Function and Its Integral (several $i$ may share a value), but grouping the disjoint intervals $(t_{i-1},t_i]$ by value and using finite additivity of $\lambda$ (from Measure, Measure Space, and Probability Measure) together with interval lengths gives exactly these sums. Pointwise

$$\varphi\ \le\ \tilde{h}\,\mathbf{1}_{(a,b]}\ \le\ \psi ,$$

since $m_i\le h\le M_i$ on $[t_{i-1},t_i]\supseteq(t_{i-1},t_i]$. Moreover $\int\tilde{h}\,\mathbf{1}_{(a,b]}\,d\lambda=\int\tilde{h}\,d\lambda$: the difference $\tilde{h}-\tilde{h}\mathbf{1}_{(a,b]}=h(a)\,\mathbf{1}_{\{a\}}$ is a simple function with integral $h(a)\,\lambda(\{a\})=0$, since the interval $[a,a]$ has length $0$; apply linearity. Hence by monotonicity (Linearity and Monotonicity of the Lebesgue Integral),

$$L(h,\mathcal{P})\ \le\ \int_{\mathbb{R}}\tilde{h}\,d\lambda\ \le\ U(h,\mathcal{P})\qquad\text{for every partition }\mathcal{P}.$$

Step 4 (Darboux–Riemann bridge and conclusion). Let $\varepsilon>0$ and choose $\delta>0$ as in Riemann Integrability on a Closed Interval for $h$ and $\varepsilon$. By the Archimedean property pick a natural number $m$ with $(b-a)/m<\delta$ and let $\mathcal{P}$ be the uniform partition $t_i=a+i(b-a)/m$, of mesh less than $\delta$. By the definition of the suprema $M_i$ there are tags $\xi_i\in[t_{i-1},t_i]$ with $h(\xi_i)>M_i-\varepsilon/(b-a)$; the corresponding Riemann sum $S^{+}=\sum_ih(\xi_i)(t_i-t_{i-1})$ satisfies $|S^{+}-I|<\varepsilon$ and

$$U(h,\mathcal{P})=\sum_iM_i(t_i-t_{i-1})\ <\ S^{+}+\frac{\varepsilon}{b-a}\sum_i(t_i-t_{i-1})\ =\ S^{+}+\varepsilon\ <\ I+2\varepsilon .$$

Symmetrically, by the definition of the infima $m_i$ there are tags $\eta_i$ with $h(\eta_i)<m_i+\varepsilon/(b-a)$, giving $L(h,\mathcal{P})>I-2\varepsilon$. Combining with Step 3 applied to this particular $\mathcal{P}$,

$$I-2\varepsilon\ <\ L(h,\mathcal{P})\ \le\ \int_{\mathbb{R}}\tilde{h}\,d\lambda\ \le\ U(h,\mathcal{P})\ <\ I+2\varepsilon .$$

Since $\varepsilon>0$ was arbitrary, $\int_{\mathbb{R}}\tilde{h}\,d\lambda=I$. Undoing the shift of Step 2 gives Claim 3 in general. $\blacksquare$