Show that the function $f : [0, 1] \to \mathbb{R}$ defined by $f(x) = cx$ for some fixed $c \in \mathbb{R}$ is Riemann Integrable

Question

Show that the function $f : [0, 1] \to \mathbb{R}$ defined by $f(x) = cx$ for some fixed $c \in \mathbb{R}$ is Riemann Integrable and that $\int_0^1 cx dx = \frac{c}{2}$

I'm trying to understand the following proof as given in my analysis course's set of notes.

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And this is Lemma 2.1 that they are referencing

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Now instead of denoting the lower and upper Darboux sums by $\underline{S}(f, P)$ and $\overline{S}(f, P)$ respectively, I'll denote them by $L(f, P)$ and $U(f, P)$ respectively.

Now as I learned it, we define $\int_{0}^{1}f(x)dx$ as the number as the number $$\int_{0}^1 f(x)dx = \inf \{U(f, P) \ | P \text{ is a partition of } [0, 1] \}= \sup \{L(f, P) \ | P \text{ is a partition of } [0, 1] \}$$ (provided this equality holds i.e. $f$ is Riemann integrable)

So to show that $f$ is Riemann integrable we need to prove that this equality holds and then to "calculate" the value of the integral we just take either the left hand side or the right hand side of this integral.

However in the beginning of the proof, I don't see why Lemma 2.1 shows that we only need to consider the partitions $P = \{x_i\}_{i=0}^n$ where $x_i = \frac{i}{n}$ for $i=0, 1, ..., n$.

If I define $P_n = \{x_i\}_{i=0}^n$ by $x_i = \frac{i}{n}$ for $i \in \{0, 1, ..., n\}$ then we can see that $P_1 = \{0, 1\}$ and $P_2 = \{0, \frac{1}{2}, 1\}$ and $P_3 = \{0, \frac{1}{3}, \frac{2}{3}, 1\}$ and generalizing we get $P_m = \{0, \frac{1}{m}, \frac{2}{m}, ...., \frac{m-1}{m}, 1\}$.

But from this we can see that $P_{k-1}$ is not necessarily a subset of $P_k$ so I don't see how we can apply Lemma 2.1 here

Moreover the set of all such partitions $\{P_n\}_{n \in \mathbb{N}}$ is only a subset of all possible partitions of $[0, 1]$

So I don't see how $\inf \{U(f, P_n) \ | P_n \in \{P_n\}_{n \in \mathbb{N}}\} = \inf \{U(f, P) \ | P \text{ is a partition of } [0, 1] \}$ and the same case for the supremums

Basically I don't see how this proof implies by the definition (or through the use of theorems/lemmas) of Riemann integrability that $f$ is Riemann integrable.

Answer 1

All we need lemma 2.1 for is

$$\tag1\inf\{\,U(P)\mid P\text{ is a partition}\,\}\ge\sup\{\,L(P)\mid P\text{ is a partition}\,\}$$ Proof. Assume otherwise. Then there exist partitions $P_1,P_2$ with $U(P_1)<L(P_2)$. But $P':=P_1\cup P_2$ is a refinement of both, hence by lemma 2.1 $L(P_2)\le L(P')\le U(P')\le U(P_1)$, contradiction. $\square$

You do not need that the equidistant partitions are refinements of one another. In fact, you could use any class of partitions (let's call them "nice" partitions), and then show

Proposition. If $$\tag2\inf\{\,U(P)\mid P\text{ is a nice partition}\,\}=\sup\{\,L(P)\mid P\text{ is a nice partition}\,\},$$ then $\inf\{\,U(P)\mid P\text{ is a partition}\,\}=\sup\{\,L(P)\mid P\text{ is a partition}\,\}$ and the integral can be computed using nice partitions.

Proof. As the nice partions are a subset of all partitions, we clearly have $$ \inf\{\,U(P)\mid P\text{ is a nice partition}\,\}\ge \inf\{\,U(P)\mid P\text{ is a partition}\,\}$$ and $$ \sup\{\,L(P)\mid P\text{ is a nice partition}\,\}\le \sup\{\,L(P)\mid P\text{ is a partition}\,\}.$$ Thus if $(2)$ holds, we conclude $$\inf\{\,U(P)\mid P\text{ is a partition}\,\}\le \sup\{\,L(P)\mid P\text{ is a partition}\,\}.$$ Then from $(1)$, we conclude that all four numbers are equal and hence by definition are equal to the integral. $\square$