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请求解析黎曼积分的两个等价定义及Apostol书中的定义7.1

Alright, let's break down the two equivalent definitions of the Riemann integral thoroughly, starting with the one referenced from Apostol's Mathematical Analysis (Definition 7.1) and pairing it with the classic upper/lower integral formulation. We'll also walk through their equivalence proof step by step.

1. The Two Equivalent Definitions of the Riemann Integral

1.1 Apostol's Definition (Uniform Convergence of Riemann Sums)

First, let's formalize the foundational concepts and complete the definition you mentioned:

Let $f$ be a bounded function on the closed interval $[a,b]$. A partition of $[a,b]$ is a finite set $P = {x_0, x_1, ..., x_n}$ where $a = x_0 < x_1 < ... < x_n = b$. For each subinterval $[x_{k-1}, x_k]$, choose any sample point $t_k \in [x_{k-1}, x_k]$. The Riemann sum of $f$ over $P$ is defined as:
$$S(P,f) = \sum_{k=1}^n f(t_k) \Delta x_k$$
where $\Delta x_k = x_k - x_{k-1}$ denotes the width of the $k$-th subinterval.

Complete Definition 1:

A function $f$ is Riemann integrable on $[a,b]$ if there exists a real number $A$ such that:
For every $\epsilon > 0$, there exists a partition $P_\epsilon$ of $[a,b]$ where, for all partitions $P$ finer than $P_\epsilon$ (i.e., $P_\epsilon \subseteq P$) and any choice of sample points $t_k \in [x_{k-1}, x_k]$, the inequality
$$|S(P,f) - A| < \epsilon$$
holds. The number $A$ is called the Riemann integral of $f$ over $[a,b]$, written as $\int_a^b f(x) dx = A$.

1.2 Classic Upper/Lower Integral Definition

This definition relies on bounds of the function over subintervals, rather than arbitrary sample points:

For any partition $P$ of $[a,b]$, define:

  • $M_k = \sup{f(x) \mid x \in [x_{k-1}, x_k]}$ (the supremum of $f$ on the $k$-th subinterval)
  • $m_k = \inf{f(x) \mid x \in [x_{k-1}, x_k]}$ (the infimum of $f$ on the $k$-th subinterval)

Then we define:

  • Upper sum: $U(P,f) = \sum_{k=1}^n M_k \Delta x_k$
  • Lower sum: $L(P,f) = \sum_{k=1}^n m_k \Delta x_k$

Using these sums, we define the upper integral and lower integral:

  • Upper integral: $\overline{\int}_a^b f(x) dx = \inf{U(P,f) \mid P \text{ is a partition of } [a,b]}$
  • Lower integral: $\underline{\int}_a^b f(x) dx = \sup{L(P,f) \mid P \text{ is a partition of } [a,b]}$

Definition 2:

A function $f$ is Riemann integrable on $[a,b]$ if its upper integral equals its lower integral:
$$\overline{\int}_a^b f(x) dx = \underline{\int}_a^b f(x) dx = A$$
Here, $A$ is the Riemann integral of $f$ over $[a,b]$.

2. Proof of Equivalence

We need to show that Definition 1 holds if and only if Definition 2 holds.

2.1 Definition 1 ⇒ Definition 2

Suppose $f$ is integrable by Definition 1, with integral $A$. Let $\epsilon > 0$ be arbitrary. By Definition 1, there exists a partition $P_\epsilon$ such that for all finer partitions $P$, every Riemann sum $S(P,f)$ satisfies $A - \epsilon < S(P,f) < A + \epsilon$.

For this partition $P$:

  • The lower sum $L(P,f)$ is the infimum of all possible Riemann sums over $P$, so $L(P,f) \geq A - \epsilon$
  • The upper sum $U(P,f)$ is the supremum of all possible Riemann sums over $P$, so $U(P,f) \leq A + \epsilon$

By definition of upper/lower integrals:
$$A - \epsilon \leq \underline{\int}_a^b f \leq \overline{\int}_a^b f \leq A + \epsilon$$
Since $\epsilon$ can be made arbitrarily small, we must have $\underline{\int}_a^b f = \overline{\int}_a^b f = A$, so $f$ satisfies Definition 2.

2.2 Definition 2 ⇒ Definition 1

Suppose $f$ is integrable by Definition 2, with $\overline{\int}_a^b f = \underline{\int}_a^b f = A$. Let $\epsilon > 0$ be arbitrary.

By definition of upper/lower integrals:

  • There exists a partition $P_1$ such that $U(P_1,f) < A + \frac{\epsilon}{2}$
  • There exists a partition $P_2$ such that $L(P_2,f) > A - \frac{\epsilon}{2}$

Let $P_\epsilon = P_1 \cup P_2$ (the common refinement of $P_1$ and $P_2$). A key property of upper/lower sums is that refining a partition does not increase the upper sum or decrease the lower sum. So for any partition $P$ finer than $P_\epsilon$:
$$A - \frac{\epsilon}{2} < L(P_2,f) \leq L(P,f) \leq S(P,f) \leq U(P,f) \leq U(P_1,f) < A + \frac{\epsilon}{2}$$
This implies $|S(P,f) - A| < \frac{\epsilon}{2} < \epsilon$, which satisfies the conditions of Definition 1.


内容的提问来源于stack exchange,提问作者user39756

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最近更新时间:2026.05.19 06:12:30