如何证明有界函数Riemann可积当且仅当Darboux可积（含陶哲轩习题）

Alright, let's dig into this foundational result connecting Riemann and Darboux integrability—this is a core topic in real analysis, so let's break it down clearly, starting with key definitions to align on terminology.

Key Definitions #

First, let's formalize the terms as stated in the problem (and Tao's Measure Theory exercise):

• Riemann Integrability: A bounded function $f:[a,b]\to\mathbb{R}$ is Riemann integrable if there exists a real number $I$ such that for every $\epsilon>0$, there exists a $\delta>0$. For any partition $\mathcal{P}$ with $|\mathcal{P}|<\delta$, all Riemann sums $\mathcal{R}(f,\mathcal{P})$ (regardless of sample point choices) satisfy $|\mathcal{R}(f,\mathcal{P}) - I|<\epsilon$. We write $I = \int_a^b f(x),dx$.
• Darboux Integrability: For a bounded $f$, define:
  • Lower Darboux integral: $\underline{\int}_a^b f = \sup\left{ \textrm{p.c.}\int_a^b g \mid g \leq f, g \text{ piecewise constant} \right}$
  • Upper Darboux integral: $\overline{\int}_a^b f = \inf\left{ \textrm{p.c.}\int_a^b h \mid h \geq f, h \text{ piecewise constant} \right}$
    $f$ is Darboux integrable if $\underline{\int}_a^b f = \overline{\int}_a^b f$, and this common value is its Darboux integral.

Proof of Equivalence #

We'll prove both directions of the "if and only if" statement to establish full equivalence.

Direction 1: Riemann Integrability $\implies$ Darboux Integrability (with equal integrals) #

Suppose $f$ is Riemann integrable with integral $I$. We need to show $\underline{\int}_a^b f = \overline{\int}_a^b f = I$.

1. Pick any $\epsilon>0$. By Riemann integrability, there exists a $\delta>0$ such that for every partition $\mathcal{P}$ with $|\mathcal{P}|<\delta$, all Riemann sums $\mathcal{R}(f,\mathcal{P})$ lie in the interval $(I-\epsilon, I+\epsilon)$.
2. Take such a partition $\mathcal{P} = {x_0=a, x_1,...,x_n=b}$. For each subinterval $[x_{i-1},x_i]$, let $m_i = \inf_{x\in[x_{i-1},x_i]} f(x)$ and $M_i = \sup_{x\in[x_{i-1},x_i]} f(x)$.
3. Construct piecewise constant functions: $g(x)=m_i$ on $[x_{i-1},x_i)$ (endpoint values don't affect the integral) and $h(x)=M_i$ on $[x_{i-1},x_i)$. Their integrals are exactly the lower Darboux sum $L(f,\mathcal{P}) = \sum_{i=1}^n m_i(x_i-x_{i-1})$ and upper Darboux sum $U(f,\mathcal{P}) = \sum_{i=1}^n M_i(x_i-x_{i-1})$.
4. Notice that $L(f,\mathcal{P}) \leq \mathcal{R}(f,\mathcal{P}) \leq U(f,\mathcal{P})$ for any sample point choice in $\mathcal{R}(f,\mathcal{P})$. Since all Riemann sums are within $\epsilon$ of $I$, we must have:
   • $L(f,\mathcal{P}) \geq I - \epsilon$ (otherwise, choosing sample points where $f$ attains $m_i$ would give a Riemann sum less than $I-\epsilon$, contradicting the definition)
   • $U(f,\mathcal{P}) \leq I + \epsilon$ (similar logic: choosing sample points where $f$ attains $M_i$ would give a sum greater than $I+\epsilon$ otherwise)
5. By definition of lower/upper Darboux integrals:
   • $\underline{\int}_a^b f \geq L(f,\mathcal{P}) \geq I - \epsilon$
   • $\overline{\int}_a^b f \leq U(f,\mathcal{P}) \leq I + \epsilon$
6. Since $\epsilon>0$ is arbitrary, we get $\underline{\int}_a^b f \geq I$ and $\overline{\int}_a^b f \leq I$. But for any bounded function, $\underline{\int}_a^b f \leq \overline{\int}_a^b f$, so the only possibility is $\underline{\int}_a^b f = \overline{\int}_a^b f = I$. Thus $f$ is Darboux integrable, with the same integral value as its Riemann integral.

Direction 2: Darboux Integrability $\implies$ Riemann Integrability (with equal integrals) #

Suppose $\underline{\int}_a^b f = \overline{\int}_a^b f = I$. We need to show $f$ is Riemann integrable with integral $I$.

1. Pick any $\epsilon>0$. By Darboux integrability, there exist piecewise constant functions $g \leq f$ and $h \geq f$ such that $\textrm{p.c.}\int_a^b h - \textrm{p.c.}\int_a^b g < \epsilon$.
2. Let $\mathcal{P}0$ be a common refinement of the partitions defining $g$ and $h$. For each subinterval $[x{i-1},x_i]$ of $\mathcal{P}_0$, let $m_i = \inf f$ and $M_i = \sup f$ on that interval. We have $g(x) \leq m_i \leq f(x) \leq M_i \leq h(x)$, so:
   $$U(f,\mathcal{P}0) - L(f,\mathcal{P}0) = \sum{i=1}^n (M_i - m_i)(x_i-x{i-1}) \leq \textrm{p.c.}\int_a^b h - \textrm{p.c.}\int_a^b g < \epsilon$$
3. Let $k$ be the number of subintervals in $\mathcal{P}0$, and let $M = \sup{[a,b]} f$, $m = \inf_{[a,b]} f$ (finite since $f$ is bounded). Choose $\delta = \frac{\epsilon}{2k(M - m)}$ (if $M=m$, $f$ is constant and trivially Riemann integrable, so we can skip this edge case).
4. Now take any partition $\mathcal{P}$ with $|\mathcal{P}|<\delta$. Let $\mathcal{P}'$ be the common refinement of $\mathcal{P}$ and $\mathcal{P}_0$. By properties of Darboux sums:
   • Refining a partition can only increase lower sums and decrease upper sums, so $U(f,\mathcal{P}) - L(f,\mathcal{P}) \leq U(f,\mathcal{P}') - L(f,\mathcal{P}')$
   • The difference $U(f,\mathcal{P}') - L(f,\mathcal{P}')$ is at most $U(f,\mathcal{P}_0) - L(f,\mathcal{P}_0) + (M - m) \cdot k \cdot \delta$. The extra term accounts for small intervals added by $\mathcal{P}$—each of the $k$ subintervals of $\mathcal{P}_0$ can be split into at most two parts by $\mathcal{P}$, so the total extra length is at most $k \cdot \delta$.
5. Substitute our choice of $\delta$:
   $$U(f,\mathcal{P}') - L(f,\mathcal{P}') < \frac{\epsilon}{2} + (M - m) \cdot k \cdot \frac{\epsilon}{2k(M - m)} = \epsilon$$
6. For any Riemann sum $\mathcal{R}(f,\mathcal{P})$, we have $L(f,\mathcal{P}) \leq \mathcal{R}(f,\mathcal{P}) \leq U(f,\mathcal{P})$, and since $L(f,\mathcal{P}) \leq I \leq U(f,\mathcal{P})$, this implies:
   $$|\mathcal{R}(f,\mathcal{P}) - I| \leq U(f,\mathcal{P}) - L(f,\mathcal{P}) < \epsilon$$
7. This satisfies the definition of Riemann integrability: for every $\epsilon>0$, we found a $\delta>0$ where all Riemann sums over partitions with $|\mathcal{P}|<\delta$ are within $\epsilon$ of $I$. Thus $f$ is Riemann integrable with integral $I$.

内容的提问来源于stack exchange，提问作者user522521