The Riemann integral · Selected prerequisite proofs

L5.1.2: Darboux sum bounds and existence of their defining finite extrema.

Proposition 5.1.2.

Let

f \colon [a,b] \to \R

be a bounded function. Let

m, M \in \R

be such that for all

x \in [a,b]\text{,}

we have

m \leq f(x) \leq M\text{.}

Then for every partition

P

of

[a,b]\text{,}

\begin{equation*} m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a) .\tag{5.1} \end{equation*}

Proof.

Let

P

be a partition of

[a,b]\text{.}

Note that for all

i\text{,}

we have

m \leq m_i \leq M_i \leq M\text{.}

We also have

\sum_{i=1}^n \Delta x_i = (b-a)\text{.}

Therefore,

\begin{gathered} m(b-a) = m \left( \sum_{i=1}^n \Delta x_i \right) = \sum_{i=1}^n m \Delta x_i \leq \sum_{i=1}^n m_i \Delta x_i \\ \leq \sum_{i=1}^n M_i \Delta x_i \leq \sum_{i=1}^n M \Delta x_i = M \left( \sum_{i=1}^n \Delta x_i \right) = M(b-a) . \end{gathered}

Hence, we get (5.1). In particular, the sets of lower and upper sums are bounded sets.

L5.1.7: Both refinement inequalities; omitted upper half supplied by P12.2.

Proposition 5.1.7.

Let

f \colon [a,b] \to \R

be a bounded function, and let

P

be a partition of

[a,b]\text{.}

Let

\widetilde{P}

be a refinement of

P\text{.}

Then

\begin{equation*} L(P,f) \leq L(\widetilde{P},f) \qquad \text{and} \qquad U(\widetilde{P},f) \leq U(P,f) . \end{equation*}

Proof.

The tricky part of this proof is to get the notation correct. Let

\widetilde{P} = \{ \widetilde{x}_0, \widetilde{x}_1, \ldots, \widetilde{x}_{\ell} \}

be a refinement of

P = \{ x_0, x_1, \ldots, x_n \}\text{.}

Then

x_0 = \widetilde{x}_0

and

x_n = \widetilde{x}_{\ell}\text{.}

In fact, there are integers

k_0 < k_1 < \cdots < k_n

such that

x_i = \widetilde{x}_{k_i}

for

i=0,1,2,\ldots,n\text{.}

Let

\Delta \widetilde{x}_q \coloneqq \widetilde{x}_q - \widetilde{x}_{q-1}

for

q=0,1,2,\ldots,\ell\text{.}

See Figure 5.2. We get

\begin{equation*} \Delta x_i = x_i - x_{i-1} = \widetilde{x}_{k_i} - \widetilde{x}_{k_{i-1}} = \sum_{q=k_{i-1}+1}^{k_i} \widetilde{x}_{q} - \widetilde{x}_{q-1} = \sum_{q=k_{i-1}+1}^{k_i} \Delta \widetilde{x}_q . \end{equation*}

A diagram of an interval between x sub quantity i minus 1 and x sub i of width Delta x sub i. It is divided into three smaller intervals. The left hand endpoint x sub quantity i minus 1 is also marked as tilde x sub quantity q minus 3 and tilde x sub quantity k sub quantity i minus 1. Similarly the right hand endpoint x sub i is also marked as tilde x sub q and tilde x sub k sub i. The two inside points are marked as tilde x sub quantity q minus 2 and tilde x sub quantity q minus 1. The three intervals are of length Delta tilde x sub quantity q minus 2, Delta tilde x sub quantity q minus 1, and Delta tilde x sub q. — Figure 5.2. Refinement of a subinterval. Notice $\Delta x_i = \Delta \widetilde{x}_{q-2} + \Delta \widetilde{x}_{q-1} + \Delta \widetilde{x}_{q}\text{,}$ and also $k_{i-1}+1 = q-2$ and $k_{i} = q\text{.}$

Let

m_i

be as before and correspond to the partition

P\text{.}

Let

\widetilde{m}_q \coloneqq \inf \bigl\{ f(x) : \widetilde{x}_{q-1} \leq x \leq \widetilde{x}_q \bigr\}\text{.}

Now,

m_i \leq \widetilde{m}_q

for

k_{i-1} < q \leq k_i\text{.}

Therefore,

\begin{equation*} m_i \Delta x_i = m_i \sum_{q=k_{i-1}+1}^{k_i} \Delta \widetilde{x}_q = \sum_{q=k_{i-1}+1}^{k_i} m_i \Delta \widetilde{x}_q \leq \sum_{q=k_{i-1}+1}^{k_i} \widetilde{m}_q \Delta \widetilde{x}_q . \end{equation*}

So

\begin{equation*} L(P,f) = \sum_{i=1}^n m_i \Delta x_i \leq \sum_{i=1}^n \, \sum_{q=k_{i-1}+1}^{k_i} \widetilde{m}_q \Delta \widetilde{x}_q = \sum_{q=1}^{\ell} \widetilde{m}_q \Delta \widetilde{x}_q = L(\widetilde{P},f). \end{equation*}

The proof of

U(\widetilde{P},f) \leq U(P,f)

is left as an exercise.

L5.1.8: Lower Darboux integral <= upper integral, with common-refinement proof.

Proposition 5.1.8.

Let

f \colon [a,b] \to \R

be a bounded function. Let

m, M \in \R

be such that for all

x \in [a,b]\text{,}

we have

m \leq f(x) \leq M\text{.}

Then

\begin{equation*} m(b-a) \leq \underline{\int_a^b} f \leq \overline{\int_a^b} f \leq M(b-a) .\tag{5.2} \end{equation*}

Proof.

By Proposition 5.1.2, for every partition

P\text{,}

\begin{equation*} m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a). \end{equation*}

The inequality

m(b-a) \leq L(P,f)

implies

m(b-a) \leq \underline{\int_a^b} f\text{.}

The inequality

U(P,f) \leq M(b-a)

implies

\overline{\int_a^b} f \leq M(b-a)\text{.}

The middle inequality in (5.2) is the main point of the proposition. Let

P_1, P_2

be partitions of

[a,b]\text{.}

Define

\widetilde{P} \coloneqq P_1 \cup P_2\text{.}

The set

\widetilde{P}

is a partition of

[a,b]\text{,}

which is a refinement of

P_1

and a refinement of

P_2\text{.}

By Proposition 5.1.7,

L(P_1,f) \leq L(\widetilde{P},f)

and

U(\widetilde{P},f) \leq U(P_2,f)\text{.}

So

\begin{equation*} L(P_1,f) \leq L(\widetilde{P},f) \leq U(\widetilde{P},f) \leq U(P_2,f) . \end{equation*}

In other words, for two arbitrary partitions

P_1

and

P_2\text{,}

we have

L(P_1,f) \leq U(P_2,f)\text{.}

Recall Proposition 1.2.7, and take the supremum and infimum over all partitions:

\begin{gathered} \underline{\int_a^b} f = \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,b] \bigr\} \\ \leq \inf \, \bigl\{ U(P,f) : P \text{ a partition of } [a,b] \bigr\} = \overline{\int_a^b} f . \qedhere \end{gathered}

L5.1.10: Integral bound is the just-proved Darboux bound with equal lower/upper integrals, as Definition 5.1.9 specifies.

Proposition 5.1.10.

Let

f \colon [a,b] \to \R

be a Riemann integrable function. Let

m, M \in \R

be such that

m \leq f(x) \leq M

for all

x \in [a,b]\text{.}

Then

\begin{equation*} m(b-a) \leq \int_a^b f \leq M(b-a) . \end{equation*}

Proof. For an integrable function the lower and upper Darboux integrals are equal to its integral. Substitute this equality in the preceding bound (5.2), proved above. This gives the two asserted inequalities.

L5.1.11: Exact constant integral, proved by coinciding upper/lower bounds.

Example 5.1.11.

We integrate constant functions using Proposition 5.1.8. If

f(x) \coloneqq c

for some constant

c\text{,}

then we take

m = M = c\text{.}

In inequality (5.2) all the inequalities must be equalities. Thus

f

is integrable on

[a,b]

and

\int_a^b f = c(b-a)\text{.}

L5.1.13: Arbitrarily small Darboux gaps imply integrability.

Proposition 5.1.13.

Let

f \colon [a,b] \to \R

be a bounded function. Then

f

is Riemann integrable if for every

\epsilon > 0\text{,}

there exists a partition

P

of

[a,b]

such that

\begin{equation*} U(P,f) - L(P,f) < \epsilon . \end{equation*}

Proof.

If for every

\epsilon > 0

such a

P

exists, then

\begin{equation*} 0 \leq \overline{\int_a^b} f - \underline{\int_a^b} f \leq U(P,f) - L(P,f) < \epsilon . \end{equation*}

Therefore,

\overline{\int_a^b} f = \underline{\int_a^b} f\text{,}

and

f

is integrable.