I'm self-studying Analysis on Manifolds by Munkres. I understood the theory of the Riemann integral over bounded rectangles and more general rectifiable sets in $\mathbb{R}^n$. In the part on improper integrals, I am becoming confused.

Munkres defines the extended or improper integral of a continuous function $f$ over an open set $A \subset \mathbf{R}^n$. He chooses any sequence $C_N$ of compact rectifiable sets such that $A = \bigcup_N C_N$ and $C_N \subset \text{Int }C_{N+1}$ for all $N$ and states that the extended integral exists if and only if the sequence $\int_{C_N} |f|$ is bounded and

$$\int_Af = \lim_{N \to \infty} \int_{C_N} f.$$

He states "...if the ordinary integral exists, then so does the extended integral and the two integrals are equal", but then " ...the extended integral may exist when the ordinary integral does not."

This makes sense if you think about integrals over intervals in $\mathbf {R}$. If a function is unbounded or the interval is unbounded then the Riemann integral does not exist but the improper integral can. But Munkres claims this even if $A$ is a bounded, open set and $f:A \to \mathbf{R}$ is a bounded, continuous function.

How is this possible?


This is true because unlike integrals over bounded intervals in $\mathbb{R}$, the nature of the boundary of $A \subset \mathbb{R}^n$ is a complicating factor for Riemann integration.

Even if $A$ is a bounded, open set and $f$ is bounded and continuous on $A$, the ordinary Riemann integral may not exist, if $A$ is not rectifiable in the sense that the measure of the boundary $\partial A$ is not zero.

The integral is defined using an enclosing rectangle $R$ as

$$\int_A f = \int_R f \chi_A,$$

and unless $f(x)$ approaches zero at essentially every boundary point from the interior, $f \chi_A$ will fail to be Riemann integrable if $M(\partial A) \neq 0$. In general, if a function is bounded and continuous inside a bounded, open set, it may not be extendable as a continuous function to the closure, e.g., $f(x) =\sin x^{-1}$ on $(0,1).$ This is never an issue for $\mathbb{R}$ since the boundary of an open interval is always a zero-measure set.

On the other hand, the extended integral exists if $\int_C f^+$ and $\int_C f^-$ are bounded for every compact rectifiable set $C \subset A$. These certainly exist when $f$ is continuous, in which case and the extended integral is

$$\int_A f = \sup_{C \subset A}\int f^+ - \sup_{C \subset A}\int f^- .$$

This will always hold for bounded continuous $f$ on a bounded open set $A$.

  • $\begingroup$ Thank you. I understand better now. One question though is you say this is never an issue for $\mathbf{R}$, but Munkres gives the example of a bounded open set $A \subset \mathbf{R}$ with non-zero measure boundary. $\endgroup$ – AlRacoon Jan 18 '18 at 17:44
  • $\begingroup$ @AlRacoon: Yes -- to exhibit a non-rectifiable bounded open set in $\mathbb{R}$ he gives that example for $n = 1$. What I meant was that for open intervals in $\mathbb{R}$ we have no problem defining the integral. Using the example $\sin x^{-1}$ we have $\int_{(0,1)} \sin x^{-1} = \int_{[0,1]} \sin x^{-1} \chi_{(0,1)}$ where the integral over the closed interval exists since the integrand is discontinuous only at the endpoint $x=0$. $\endgroup$ – RRL Jan 18 '18 at 18:56
  • $\begingroup$ FYI - some posts for related topics in this section of Munkres: here and here $\endgroup$ – RRL Jan 18 '18 at 19:02
  • $\begingroup$ Also this question shows how it is possible to ignore the contributions from sub-rectangles in a partition that intersect $\partial A$ in evaluating the integral $\int_A f$ as a limit of Riemann sums when the partition norm approaches zero. $\endgroup$ – RRL Jan 18 '18 at 20:50

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