Chapter 07 The Riemann Integral

7.4 Properties of the Integral

7.4.2.

Assume $f$ and $g$ are integrable functions on the interval $[a,b]$.

(i) The function $f + g$ is integrable on $[a,b]$ with $\int_{a}^{b} (f+g) = \int_{a}^{b}f + \int_{a}^{b} g$

Proof: Given any $\epsilon$, we can find partition $P$

$$\int_{a}^{b} (f+g) \leq U(f+g, P) = U(f, P) + U(g, P) < L(f, P) + L(g, P) + \epsilon \\ < \int_{a}^{b} f + \int_{a}^{b} g + \epsilon$$

On the other hand,

$$\int_{a}^{b} f + \int_{a}^{b} g \leq U(f, P) + U(g, P) < L(f, P) + L(g, P) + \epsilon = L(f+g, P) + \epsilon \\ \leq \int_{a}^{b} (f+g) + \epsilon$$

So we have $\int_{a}^{b} (f+g) = \int_{a}^{b}f + \int_{a}^{b} g$.

$\square$

(ii) For $k ∈ \mathbf{R}$, the function $kf$ is integrable with $\int_{a}^{b} kf = k \int_{a}^{b} f$.

Proof:

$$\int_{a}^{b} kf \leq U(kf, P) = k U(f, P) < k L(f, P) + \epsilon \leq k \int_{a}^{b} f$$

On the other hand

$$k \int_{a}^{b} f \leq k U(f, P) < k L(f, P) + \epsilon = L(kf, P) + \epsilon \leq \int_{a}^{b} kf + \epsilon$$

$\square$

7.5 The Fundamental Theorem of Calculus

Theorem 7.5.1 (Fundamental Theorem of Calculus).

(i) If $f : [a,b] \rightarrow \mathbf{R}$ is integrable, and $F : [a,b] \rightarrow \mathbf{R}$ satisfies $F'(x) = f(x)$ for all $x \in [a, b]$, then

$$\int_{a}^{b} f = F(b) - F(a).$$

$\square$

(ii) Let $g:[a, b] \rightarrow \mathbf{R}$ be integrable, and for $x \in [a, b]$, define

$$G(x) = \int_{a}^{x} g(x)$$

Then $G$ is continuous on $[a,b]$. If $g$ is continuous at some point $c ∈ [a,b]$, then $G$ is differentiable at $c$ and $G'(c) = g(c)$ .

Proof:

(i) Given a partition $P, a = x_0 < x_1 < \cdots < x_n = b$.

$$F(x_{k+1}) - F(x_k) = f(x'_k) (x_{k+1} - x_k)$$

Since $m_k \leq f(x'_k) \leq M_k$, then $L(f, P) \leq F(b) - F(a) \leq U(f, P)$,

So $\int_{a}^{b} f = F(b) - F(a)$.

(ii)

$$|G(y) - G(x)| = \left| \int_{x}^{y} g(t) \right| \leq \int_{x}^{y} |g(t)| \\ \leq M(x - y)$$

$G$ is Lipschitz and so is uniformly continuous on $[a,b]$.

For the second part:

$$\left| \frac{ G(x) - G(c) }{ x - c } - g(c) \right| = \frac{1}{x - c} \left| \int_{c}^{x} g(t) - g(c) (x - c) \right| \\ = \frac{1}{x - c} \left| \int_{c}^{x} (g(t) - g(c)) \right| \\ \leq \frac{1}{x - c} \int_{c}^{x} \left| (g(t) - g(c)) \right| \\ \leq \frac{1}{x - c} \epsilon (x - c) = \epsilon.$$

$\square$