Suppose $(a,b)$ is an open and bounded interval of $\mathbb{R}$. Also, let $$f:(a,b)\to \mathbb{R}$$ be a bounded function. Set $\tilde{f}:[a,b]\to\mathbb{R}$ as the function $$\tilde{f}(x)=\begin{cases} f(x)\qquad x \in (a,b)\\\alpha \qquad x=a\\\beta \qquad x=b\end{cases}.$$ Is it true that if $\tilde{f}$ is Riemann integrable over $[a,b]$ then the improper integral of $f$ over $(a,b)$ converge and $$\int_{a}^{b}f=\int_{a}^{b}\tilde{f}\qquad ?$$

Note that the improper integral of $f$ over $(a,b)$, provided that $f$ is riemann integrable on any compact subinterval $[c,d]\subset (a,b)$, is defined as $$\int_{a}^{b}f=\lim_{m\to a^+}\int_{m}^{c}f+\lim_{M\to b^-}\int_{c}^{M}f$$ where $c\in (a,b)$ is arbitrary.

Personal attempt Since $\tilde{f}$ is riemann integrable over $[a,b]$ then it's riemann integrable over any $[c,d]\subset (a,b)$. But $\tilde{f}=f$ on such compact intervals. So $f$ is riemann integrable over any $[c,d]\subset (a,b)$. Then, for any fixed $c\in (a,b)$ one has $$\int_{a}^{b}f(t)dt=\lim_{m\to a^+}\int_{m}^{c}f(t)dt+\lim_{M\to b^-}\int_{c}^{M}f(t)dt=\lim_{m\to a^+}\int_{m}^{c}\tilde{f}(t)dt+\lim_{M\to b^-}\int_{c}^{M}\tilde{f}(t)dt=\int_{a}^{c}\tilde{f}+\int_{c}^{b}\tilde{f}=\int_{a}^{b}\tilde{f}.$$


Your proof is correct except for the fact that you need to justify the last step (the use of second last '$=$' sign). It is based on the Fundamental Theorem of Calculus a part of which says that if a function $f:[a, b] \to\mathbb {R} $ is Riemann integrable on $[a, b] $ and if $F(x) =\int_{a} ^{x} f(t) \, dt$ then the function $F$ is continuous on $[a, b] $.

This also shows that the notion of improper integral for bounded functions on bounded intervals is superfluous (but perhaps attractive to many as some questions on this website suggest). The usual (or shall we say proper) Riemann integral is powerful enough to deal with bounded functions on bounded intervals and this is a precondition for the definition of Riemann integral. For example the integral $\int_{0}^{1}\sin(1/x)\,dx$ is not improper but $\int_{0}^{1}(1-x^2)^{-1/2}\,dx$ is improper (and important at the same time).

| cite | improve this answer | |
  • $\begingroup$ yes, sir. I forgot to write it. So, if I well understood, the same reasoning applies to any bounded function $f:[a,b]\setminus\{c_1,\dots,c_n\}\to\mathbb{R}$ that is NOT defined at a finite number of points $c_1,\dots,c_n$ within $[a,b]$. If $\tilde{f}$ is a new function such that $\tilde{f}=f$ on $[a,b]\setminus\{c_1,\dots,c_n\}$ and anything else on $\{c_1,\dots,c_n\}$ then if $\tilde{f}$ is Riemann integrable on $[a,b]$ also the improper integral of $f$ on $[a,b]$ converge. Right? $\endgroup$ – eleguitar Nov 25 '18 at 16:57
  • $\begingroup$ @eleguitar: I view such integrals as proper. There is nothing extra in defining them via limit of an integral. If your function is undefined at a finite number of point then define them in any manner at those points and then integrate the resulting based. But in general one can't modify a function at an infinite number of points without affecting its Riemann integrability or the value of its integral. $\endgroup$ – Paramanand Singh Nov 25 '18 at 16:59
  • $\begingroup$ Perfect, thanks. Just last question: is it the same if I replace $\{c_1,\dots,c_n\}$ with any $E\subset[a,b]$ which has Lebesgue measure zero? I think the answer is no. I would like to extend Lebesgue's criterion for Riemann Integrability to function that are not necessarily defined on every point of $[a,b]$. $\endgroup$ – eleguitar Nov 25 '18 at 17:20
  • $\begingroup$ @eleguitar: the measure zero sets are quite different beasts. In terms of cardinality they can be uncountable like Cantor set. To use Lebesgue criterion the right approach is to ensure a definition of the function on full interval $[a, b] $ and then check the set of discontinuities. Why do you want to deal with integrating functions which are undefined at some points? $\endgroup$ – Paramanand Singh Nov 25 '18 at 17:25
  • $\begingroup$ @eleguitar: just to let you know, the improper integrals were invented not to deal with functions undefined at certain points, but rather to deal with unbounded functions and / or unbounded intervals. $\endgroup$ – Paramanand Singh Nov 25 '18 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.