1.2: Preparation

We begin with two useful convergence criteria for improper integrals that do not involve a parameter. Consistent with the definition on p. 152, we say that $f$ is locally integrable on an interval $I$ if it is integrable on every finite closed subinterval of $I$.

[theorem:2] Suppose $g$ is locally integrable on $[a,b)$ and denote

$$G(r)=\int_{a}^{r}g(x)\,dx,\quad a\le r<b.$$

Then the improper integral $\int_{a}^{b}g(x)\,dx$ converges if and only if$,$ for each $\epsilon >0,$ there is an $r_{0}\in[a,b)$ such that

$$\label{eq:9} |G(r)-G(r_{1})|<\epsilon,\quad r_{0}\le r,r_{1}<b.$$

For necessity, suppose $\int_{a}^{b}g(x)\,dx=L$. By definition, this means that for each $\epsilon>0$ there is an $r_{0}\in [a,b)$ such that

$$|G(r)-L|<\frac{\epsilon}{2} \text{\quad and\quad} |G(r_{1})-L|<\frac{\epsilon}{2},\quad r_{0}\le r,r_{1}<b.$$

Therefore

$$\begin{aligned} |G(r)-G(r_{1})|&=&|(G(r)-L)-(G(r_{1})-L)|\\ &\le& |G(r)-L|+|G(r_{1})-L|< \epsilon,\quad r_{0}\le r,r_{1}<b.\end{aligned}$$

For sufficiency, [eq:9] implies that

$$|G(r)|= |G(r_{1})+(G(r)-G(r_{1}))|< |G(r_{1})|+|G(r)-G(r_{1})|\le |G(r_{1})|+\epsilon,$$

$r_{0}\le r\le r_{1}<b$. Since $G$ is also bounded on the compact set $[a,r_{0}]$ (Theorem 5.2.11, p. 313), $G$ is bounded on $[a,b)$. Therefore the monotonic functions

$$\overline{G}(r)=\sup\left\{G(r_{1})\, \big|\, r\le r_{1}<b\right\} \text{\quad and\quad} \underline{G}(r)=\inf\left\{G(r_{1})\, \big|\, r\le r_{1}<b\right\}$$

are well defined on $[a,b)$, and

$$\lim_{r\to b-}\overline{G}(r)=\overline{L} \text{\quad and\quad} \lim_{r\to b-}\underline{G}(r)=\underline{L}$$

both exist and are finite (Theorem 2.1.11, p. 47). From [eq:9],

$$\begin{aligned} |G(r)-G(r_{1})|&=&|(G(r)-G(r_{0}))-(G(r_{1})-G(r_{0}))|\\ &\le &|G(r)-G(r_{0})|+|G(r_{1})-G(r_{0})|< 2\epsilon,\end{aligned}$$

so

$$\overline{G}(r)-\underline{G}(r)\le 2\epsilon, \quad r_{0}\le r, r_{1}<b.$$

Since $\epsilon$ is an arbitrary positive number, this implies that

$$\lim_{r\to b-}(\overline{G}(r)-\underline{G}(r))=0,$$

so $\overline{L}=\underline{L}$. Let $L=\overline{L}=\underline{L}$. Since

$$\underline{G}(r)\le G(r)\le \overline{G}(r),$$

it follows that $\lim_{r\to b-} G(r)=L$.

We leave the proof of the following theorem to you (Exercise [exer:2]).

[theorem:3] Suppose $g$ is locally integrable on $(a,b]$ and denote

$$G(r)=\int_{r}^{b}g(x)\,dx,\quad a\le r<b.$$

Then the improper integral $\int_{a}^{b}g(x)\,dx$ converges if and only if$,$ for each $\epsilon >0,$ there is an $r_{0}\in(a,b]$ such that

$$|G(r)-G(r_{1})|<\epsilon,\quad a<r,r_{1}\le r_{0}.$$

To see why we associate Theorems [theorem:2] and [theorem:3] with Cauchy, compare them with Theorem 4.3.5 (p. 204)