Integration Theorems

Proposition. A continuous function $f$ on $[a, b]$ is Riemann integrable.

Proof.

Since $f$ is continuous, it is uniformly continuous.

Suppose that $\varepsilon > 0$. There exists $\delta$ such that if $\vert x_1 - x_2 \vert < \delta$, $\vert f(x_1) - f(x_2) \vert < \varepsilon / (b - a)$. Thus, any dissection $D$ with subintervals $\Set{I_1, …, I_k}$ and $\delta^\ast(D_n) < \delta$, for each subinterval $I_r$,

\[M_{I_r} - m_{I_r} < {\varepsilon \over b - a}\]

Hence,

\[S_D - s_D = \sum_{r = 1}^k (M_{I_r} - m_{I_r}) \delta_r < {\varepsilon \over b - a} \sum_{r=1}^k \delta_r = \varepsilon\]

Proposition. A monotonic function $f$ on $[a, b]$ is Riemann integrable.

Proof.

wlog. Suppose $f$ is increasing.

Suppose that $\varepsilon > 0$. Choose $N$ so that $N > (f(b) - f(a))(b - a)/\varepsilon$. Let $D$ be a dissection with $N$ intervals of equal length, i.e. $\delta_r = (b - a) / N$. Then $M_{I_r} = f(x_r)$ and $m_{I_r} \ge f(x_{r-1})$. Hence,

\[S_D - s_D \le \sum_{r=1}^N \left( f(x_r) - f(x_{r-1}) \right) {(b - a) \over N} = {(f(b) - f(a))(b - a) \over N} < \varepsilon\]

Proposition. A bounded function $f$ that is continuous on $(a, b)$ is Riemann integrable on $[a, b]$.

Proof.

Suppose that $\varepsilon > 0$ and $\vert f(x) \vert \le C$ for every $x \in [a, b]$. Choose $a’$ such that $a’ - a < \varepsilon / 8C$ and $b’$ such that $b - b’ < \varepsilon/8C$. Since $f$ is continuous on $[a’, b’]$, it is integrable and we can find a dissection $D’$ with points $x_1 = a’ < … < x_{n-1} = b’$ such that $S_{D’} - s_{D’} < \varepsilon/2$.

Let $D$ be the dissection with points $a = x_0 < x_1 < … < x_{n-1} < x_n = b$. Hence,

\[S_D - s_D \le (C - (-C)) {\varepsilon \over 8C} + (S_{D'} - s_{D'}) + (C - (-C)) {\varepsilon \over 8C} < \varepsilon\]

It means function needs not be continuous at the endpoint so to be integrable, an example will be $\int_0^1 \sin 1/x \,dx$.

References