-1
$\begingroup$

Can somebody please explain , in simple terms , if continuity ALWAYS imply integrability ? If it doesn't ? Maybe some counter examples?

Or maybe even what are the necessary conditions in order to imply that a continuous function can be integrable?

Moreover, i wanted to be sure that differentiabilility always implies continuity ... is this correct? I mean if i say that f is differentiable on point a then f is continuous on point a . is that wrong to say ? Or may it be right to say that if a left and right derivative of a fucntion exist at a point a , then there exists the left and right continuity at point a. Left and right continuity at point a together imply continuity .

$\endgroup$
  • 1
    $\begingroup$ The answer depends on what type of integrals you are considering. If you are talking about Riemann integrability on a closed interval $[a,b]$ then any continuous function is integrable. If you are talking about improper Riemann integrals or Lebesgue integrals continuity does not imply integrability. $\endgroup$ – Kavi Rama Murthy Jun 19 '18 at 7:32
  • $\begingroup$ It depends on the domain too. If the domain isn't compact the integral might not exist. Consider integrating $f(x) = x$ over $\mathbb{R}$. $\endgroup$ – Shervin Sorouri Jun 19 '18 at 7:34
  • $\begingroup$ Sorry there is a mistake in my comment. $f(x) = x$ is integrable over $\mathbb{R}$ and the value is $0$. Change the domain to $\mathbb{R}^{+}$. $\endgroup$ – Shervin Sorouri Jun 19 '18 at 7:54
  • $\begingroup$ @ShervinSorouri When I said improper integrals I included integrals over infinite intervals. So your comment is already part of my comment. $\endgroup$ – Kavi Rama Murthy Jun 19 '18 at 8:31
  • $\begingroup$ Yeah but at the time that you had commented i was writing mine. 2 mins late :( $\endgroup$ – Shervin Sorouri Jun 19 '18 at 10:32
0
$\begingroup$

This answer refers to single variable Riemann integrability. It is easier to define integrability for bounded functions and due to Weierstrass' theorem, a continuous function on a closed interval is bounded. Indeed any continuous function on a closed interval is integrable (but not any bounded function on a closed interval: for example, Dirichlet function = indicator of rational numbers, isn't integrable). However, not any continuous function on an open interval is integrable; For example take $1/x$ in $(0,1)$.

$\endgroup$
0
$\begingroup$

I found this, slightly technical, definition, which should be relevant:

What is Riemann integrable function?

A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure)

Secondly, differentiability implies continuity at a point. See this

$\endgroup$

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.