Properties of the integral · Selected prerequisite proofs

L5.2.1: Additivity of upper and lower integrals on adjacent intervals.

Lemma 5.2.1.

Suppose

a < b < c

and

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

is a bounded function. Then

\begin{equation*} \underline{\int_a^c} f = \underline{\int_a^b} f + \underline{\int_b^c} f \quad \text{and} \quad \overline{\int_a^c} f = \overline{\int_a^b} f + \overline{\int_b^c} f . \end{equation*}

Proof.

If we have partitions

P_1 = \{ x_0,x_1,\ldots,x_k \}

of

[a,b]

and

P_2 = \{ x_k, x_{k+1}, \ldots, x_n \}

of

[b,c]\text{,}

then the set

P \coloneqq P_1 \cup P_2 = \{ x_0, x_1, \ldots, x_n \}

is a partition of

[a,c]\text{.}

We find

\begin{equation*} L(P,f) = \sum_{i=1}^n m_i \Delta x_i = \sum_{i=1}^k m_i \Delta x_i + \sum_{i=k+1}^n m_i \Delta x_i = L(P_1,f) + L(P_2,f) . \end{equation*}

When we take the supremum of the right-hand side over all

P_1

and

P_2\text{,}

we are taking a supremum of the left-hand side over all partitions

P

of

[a,c]

that contain

b\text{.}

If

Q

is a partition of

[a,c]

and

P = Q \cup \{ b \}\text{,}

then

P

is a refinement of

Q

and so

L(Q,f) \leq L(P,f)\text{.}

Therefore, taking a supremum only over the

P

that contain

b

is sufficient to find the supremum of

L(P,f)

over all partitions

P\text{,}

see Exercise 1.1.9. Finally, recall Exercise 1.2.9 to compute

\begin{equation*} \begin{split} \underline{\int_a^c} f & = \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,c] \bigr\} \\ & = \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,c], b \in P \bigr\} \\ & = \sup \, \bigl\{ L(P_1,f) + L(P_2,f) : P_1 \text{ a partition of } [a,b], P_2 \text{ a partition of } [b,c] \bigr\} \\ & = \sup \, \bigl\{ L(P_1,f) : P_1 \text{ a partition of } [a,b] \bigr\} + \sup \, \bigl\{ L(P_2,f) : P_2 \text{ a partition of } [b,c] \bigr\} \\ &= \underline{\int_a^b} f + \underline{\int_b^c} f . \end{split} \end{equation*}

Similarly, for

P\text{,}

P_1\text{,}

and

P_2

as above, we obtain

\begin{equation*} U(P,f) = \sum_{i=1}^n M_i \Delta x_i = \sum_{i=1}^k M_i \Delta x_i + \sum_{i=k+1}^n M_i \Delta x_i = U(P_1,f) + U(P_2,f) . \end{equation*}

We wish to take the infimum on the right over all

P_1

and

P_2\text{,}

and so we are taking the infimum over all partitions

P

of

[a,c]

that contain

b\text{.}

If

Q

is a partition of

[a,c]

and

P = Q \cup \{ b \}\text{,}

then

P

is a refinement of

Q

and so

U(Q,f) \geq U(P,f)\text{.}

Therefore, taking an infimum only over the

P

that contain

b

is sufficient to find the infimum of

U(P,f)

for all

P\text{.}

We obtain

\begin{equation*} \overline{\int_a^c} f = \overline{\int_a^b} f + \overline{\int_b^c} f . \qedhere \end{equation*}

L5.2.2: Integrability iff both interval restrictions are integrable; interval additivity.

Proposition 5.2.2.

Let

a < b < c\text{.}

A function

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

is Riemann integrable if and only if

f

is Riemann integrable on

[a,b]

and

[b,c]\text{.}

If

f

is Riemann integrable, then

\begin{equation*} \int_a^c f = \int_a^b f + \int_b^c f . \end{equation*}

Proof.

Suppose

f \in \sR\bigl([a,c]\bigr)\text{.}

Then it is bounded and

\overline{\int_a^c} f = \underline{\int_a^c} f = \int_a^c f\text{.}

The lemma gives

\begin{equation*} \int_a^c f = \underline{\int_a^c} f = \underline{\int_a^b} f + \underline{\int_b^c} f \leq \overline{\int_a^b} f + \overline{\int_b^c} f = \overline{\int_a^c} f = \int_a^c f . \end{equation*}

Thus the inequality is an equality:

\begin{equation*} \underline{\int_a^b} f + \underline{\int_b^c} f = \overline{\int_a^b} f + \overline{\int_b^c} f . \end{equation*}

As we also know

\underline{\int_a^b} f \leq \overline{\int_a^b} f

and

\underline{\int_b^c} f \leq \overline{\int_b^c} f\text{,}

we conclude

\begin{equation*} \underline{\int_a^b} f = \overline{\int_a^b} f \qquad \text{and} \qquad \underline{\int_b^c} f = \overline{\int_b^c} f . \end{equation*}

Thus

f

is Riemann integrable on

[a,b]

and

[b,c]

and the desired formula holds.

Now assume

f

is Riemann integrable on

[a,b]

and on

[b,c]\text{.}

Again it is bounded, and the lemma gives

\begin{equation*} \underline{\int_a^c} f = \underline{\int_a^b} f + \underline{\int_b^c} f = \int_a^b f + \int_b^c f = \overline{\int_a^b} f + \overline{\int_b^c} f = \overline{\int_a^c} f . \end{equation*}

Therefore,

f

is Riemann integrable on

[a,c]\text{,}

and the integral is computed as indicated.

L5.2.4-positive: The actually written nonnegative-scalar proof only; omitted negative and sum cases are supplied locally.

Proposition 5.2.4. Linearity.

Let

f

and

g

be in

\sR\bigl([a,b]\bigr)

and

\alpha \in \R\text{.}

$\alpha f$ is in $\sR\bigl([a,b]\bigr)$ and

$\begin{equation*} \int_a^b \alpha f(x) \,dx = \alpha \int_a^b f(x) \,dx . \end{equation*}$
$f+g$ is in $\sR\bigl([a,b]\bigr)$ and

$\begin{equation*} \int_a^b \bigl( f(x)+g(x) \bigr) \,dx = \int_a^b f(x) \,dx + \int_a^b g(x) \,dx . \end{equation*}$

Proof.

Let us prove the first item for

\alpha \geq 0\text{.}

Let

P

be a partition of

[a,b]\text{,}

and

m_i \coloneqq \inf \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\}

as usual. As

\alpha \geq 0\text{,}

the multiplication by

\alpha

moves past the infimum,

\begin{equation*} \inf \bigl\{ \alpha f(x) : x \in [x_{i-1},x_i] \bigr\} = \alpha \inf \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\} = \alpha m_i . \end{equation*}

Therefore,

\begin{equation*} L(P,\alpha f) = \sum_{i=1}^n \alpha m_i \Delta x_i = \alpha \sum_{i=1}^n m_i \Delta x_i = \alpha L(P,f). \end{equation*}

Similarly,

\begin{equation*} U(P,\alpha f) = \alpha U(P,f) . \end{equation*}

Again, as

\alpha \geq 0\text{,}

we may move multiplication by

\alpha

past the supremum. Hence,

\begin{equation*} \begin{split} \underline{\int_a^b} \alpha f(x)\,dx & = \sup \, \bigl\{ L(P,\alpha f) : P \text{ a partition of } [a,b] \bigr\} \\ & = \sup \, \bigl\{ \alpha L(P,f) : P \text{ a partition of } [a,b] \bigr\} \\ & = \alpha \, \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,b] \bigr\} \\ & = \alpha \underline{\int_a^b} f(x)\,dx . \end{split} \end{equation*}

Similarly, we show

\begin{equation*} \overline{\int_a^b} \alpha f(x)\,dx = \alpha \overline{\int_a^b} f(x)\,dx . \end{equation*}

The conclusion now follows for

\alpha \geq 0\text{.}

To finish the proof of the first item (for

\alpha < 0

), we need to show that

-f

is Riemann integrable and

\int_a^b - f(x)\,dx = - \int_a^b f(x)\,dx\text{.}

The proof of this fact is left as Exercise 5.2.1.

The proof of the second item is left as Exercise 5.2.2. It is not difficult, but it is not as trivial as it may appear at first glance.

L5.2.6: Integral monotonicity via extrema and finite sums.

Proposition 5.2.6. Monotonicity.

Let

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

and

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

be bounded, and

f(x) \leq g(x)

for all

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

Then

\begin{equation*} \underline{\int_a^b} f \leq \underline{\int_a^b} g \qquad \text{and} \qquad \overline{\int_a^b} f \leq \overline{\int_a^b} g . \end{equation*}

Moreover, if

f

and

g

are in

\sR\bigl([a,b]\bigr)\text{,}

then

\begin{equation*} \int_a^b f \leq \int_a^b g . \end{equation*}

Proof.

Let

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

be a partition of

[a,b]\text{.}

Then let

\begin{equation*} m_i \coloneqq \inf \, \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\} \qquad \text{and} \qquad \widetilde{m}_i \coloneqq \inf \, \bigl\{ g(x) : x \in [x_{i-1},x_i] \bigr\} . \end{equation*}

As

f(x) \leq g(x)\text{,}

we have

m_i \leq \widetilde{m}_i\text{.}

Therefore,

\begin{equation*} L(P,f) = \sum_{i=1}^n m_i \Delta x_i \leq \sum_{i=1}^n \widetilde{m}_i \Delta x_i = L(P,g) . \end{equation*}

We take the supremum over all

P

(see Proposition 1.3.7) to obtain

\begin{equation*} \underline{\int_a^b} f \leq \underline{\int_a^b} g . \end{equation*}

Similarly, we obtain the same conclusion for the upper integrals. Finally, if

f

and

g

are Riemann integrable all the integrals are equal, and the conclusion follows.

L5.2.7: Continuous functions on compact intervals are Riemann integrable.

Lemma 5.2.7.

If

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

is a continuous function, then

f \in \sR\bigl([a,b]\bigr)\text{.}

Proof.

As

f

is continuous on a closed bounded interval, it is bounded and uniformly continuous. Given

\epsilon > 0\text{,}

find a

\delta > 0

such that

\sabs{x-y} < \delta

implies

\babs{f(x)-f(y)} < \frac{\epsilon}{b-a}\text{.}

Let

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

be a partition of

[a,b]

such that

\Delta x_i < \delta

for all

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

For example, take

n

such that

\frac{b-a}{n} < \delta\text{,}

and let

x_i \coloneqq \frac{i}{n}(b-a) + a\text{.}

Then for all

x, y \in [x_{i-1},x_i]\text{,}

we have

\sabs{x-y} \leq \Delta x_i < \delta\text{,}

and so

\begin{equation*} f(x)-f(y) \leq \babs{f(x)-f(y)} < \frac{\epsilon}{b-a} . \end{equation*}

As

f

is continuous on

[x_{i-1},x_i]\text{,}

it attains a maximum and a minimum on this interval. Let

x

be a point where

f

attains the maximum and

y

be a point where

f

attains the minimum. Then

f(x) = M_i

and

f(y) = m_i

in the notation from the definition of the integral. Therefore,

\begin{equation*} M_i-m_i = f(x)-f(y) < \frac{\epsilon}{b-a} . \end{equation*}

And so

\begin{equation*} \begin{split} \overline{\int_a^b} f - \underline{\int_a^b} f & \leq U(P,f) - L(P,f) \\ & = \left( \sum_{i=1}^n M_i \Delta x_i \right) - \left( \sum_{i=1}^n m_i \Delta x_i \right) \\ & = \sum_{i=1}^n (M_i-m_i) \Delta x_i \\ & < \frac{\epsilon}{b-a} \sum_{i=1}^n \Delta x_i \\ & = \frac{\epsilon}{b-a} (b-a) = \epsilon . \end{split} \end{equation*}

As

\epsilon > 0

was arbitrary,

\begin{equation*} \overline{\int_a^b} f = \underline{\int_a^b} f , \end{equation*}

and

f

is Riemann integrable on

[a,b]\text{.}