Let's break this proof down step by step, sticking strictly to the definitions you've laid out. First, let's recap all the key terminology to avoid confusion:

Key Definitions #

• A partition $P$ of $[a,b]$ is a finite set of points containing $a$ and $b$, usually written as $P = {a = t_0, t_1, ..., t_n = b}$ where $t_0 < t_1 < ... < t_n$. The set of all such partitions is denoted $P_{[a,b]}$.
• For each subinterval $[t_{i-1}, t_i]$ of partition $P$:
  • $M_i = \sup{f(x) : x \in [t_{i-1}, t_i]}$ (the supremum of $f$ on the subinterval)
  • $m_i = \inf{f(x) : x \in [t_{i-1}, t_i]}$ (the infimum of $f$ on the subinterval)
• The upper sum of $f$ over partition $P$:
  $$\overline{S}(f,P) = \sum_{i=1}^n M_i(t_i - t_{i-1})$$
• The upper integral of $f$ on $[a,b]$:
  $$\overline{\int}a^b f(x)dx = \inf\left{ \overline{S}(f,P) : P \in P{[a,b]} \right}$$

Our goal is to prove: If $f$ is Riemann integrable on $[a,b]$ (meaning its upper integral equals its lower integral, denoted $\int_a^b f(x)dx$), then the limit of the upper sums as the partition's mesh size tends to 0 equals the Riemann integral:
$$\lim_{|P| \to 0} \overline{S}(f,P) = \int_a^b f(x)dx$$

Proof Step 1: Use the definition of the upper integral to find a "good" partition #

By the definition of infimum, for any $\epsilon > 0$, there exists some partition $P_0 \in P_{[a,b]}$ such that:
$$\overline{S}(f,P_0) < \overline{\int}_a^b f(x)dx + \frac{\epsilon}{2}$$

Let $k$ be the number of points in $P_0$. Since $f$ is bounded on $[a,b]$, let $M = \sup{f(x) : x \in [a,b]}$ and $m = \inf{f(x) : x \in [a,b]}$ (so $M - m$ is a positive finite number). Define:
$$\delta = \frac{\epsilon}{2k(M - m)}$$

Proof Step 2: Compare upper sums of fine partitions to the "good" partition's upper sum #

Take any partition $P$ with mesh size $|P| = \max(t_i - t_{i-1}) < \delta$. Let $P' = P \cup P_0$ (this is a refinement of $P$, since we've added all points from $P_0$ to $P$).

For each subinterval $[t_{i-1}, t_i]$ of $P$:

• If it doesn't contain any points from $P_0$, its contribution to the upper sum stays the same in $P'$.
• If it does contain points from $P_0$, it gets split into at most $k$ smaller subintervals. For these split subintervals, the sum of their upper contributions is at most $M_i(t_i - t_{i-1})$ (since splitting a subinterval can't increase the supremum on smaller pieces).

The total difference between $\overline{S}(f,P)$ and $\overline{S}(f,P')$ is bounded by:
$$\overline{S}(f,P) - \overline{S}(f,P') \leq (M - m) \times k \times \delta$$

Substitute our definition of $\delta$:
$$(M - m) \times k \times \frac{\epsilon}{2k(M - m)} = \frac{\epsilon}{2}$$

So we get:
$$\overline{S}(f,P) \leq \overline{S}(f,P') + \frac{\epsilon}{2}$$

Proof Step 3: Bound the upper sum relative to the upper integral #

Since $P'$ is a refinement of $P_0$, upper sums decrease (or stay the same) when we refine a partition (smaller subintervals can't have larger suprema). So:
$$\overline{S}(f,P') \leq \overline{S}(f,P_0)$$

Combine this with our earlier result from Step 1:
$$\overline{S}(f,P) \leq \overline{S}(f,P_0) + \frac{\epsilon}{2} < \overline{\int}_a^b f(x)dx + \frac{\epsilon}{2} + \frac{\epsilon}{2} = \overline{\int}_a^b f(x)dx + \epsilon$$

Also, by definition of the upper integral, every upper sum is at least the upper integral:
$$\overline{S}(f,P) \geq \overline{\int}_a^b f(x)dx$$

Putting these together, for all partitions $P$ with $|P| < \delta$:
$$\overline{\int}_a^b f(x)dx \leq \overline{S}(f,P) < \overline{\int}_a^b f(x)dx + \epsilon$$

Proof Step 4: Finalize using Riemann integrability #

If $f$ is Riemann integrable, its upper integral equals its lower integral, which is exactly the Riemann integral:
$$\overline{\int}_a^b f(x)dx = \underline{\int}_a^b f(x)dx = \int_a^b f(x)dx$$

So the inequality above becomes:
$$\left| \overline{S}(f,P) - \int_a^b f(x)dx \right| < \epsilon$$

By the definition of the limit, this means:
$$\lim_{|P| \to 0} \overline{S}(f,P) = \int_a^b f(x)dx$$

内容的提问来源于stack exchange，提问作者Marko A. Rodríguez