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I have been doing lots of calculus these days and i want to confirm with you guys my understanding of an important concept of calculus.

Basically, in the initial phase,students assume that integration and differentiation are always associated to each other, i.e., a function which is integrable is also differentiable at the same time. But having explored on it more, i found out its not true at all and does not hold always. Many a times (or should i say infinitely times) a function can be integrable on an interval while its not differentiable on that same interval (and vice versa) .

What i want to ask is this : i recently read a conclusion on the above mentioned concept which is :

{Differentiable functions} $\subset$ {Continuous functions} $\subset$ {Integrable functions}

i.e., each is a proper subset of the next. Now, "Differentiable functions set" is a proper subset of "Continuous functions set"... that is very well understood without a doubt as every continuous function may or may not be differentiable. I have problem with the next relation which is :

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"Continuous functions set" is a proper subset of "Integrable functions set"...Why is this so???

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I am just not able to visualize this. I know that a bounded continuous function on a closed interval is integrable, well and fine, but there are unbounded continuous functions too with domain R , which we cant say will be integrable or not.

So, my question is simple. Why are there more number of elements in the "Integrable functions set" than "Continuous functions set" (here by elements i mean integrable and continuous functions ofcourse) ???. So,this is it... Can anyone plz help me understand this out in as simple words as possible. I know i need some kind of visualization which i guess is easy, but i could not make it out on my own, so i turned to u guys.Thanks for any help.

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  • $\begingroup$ or in simple words : Why are there always more integrable functions present than continuous functions?? $\endgroup$ – under root Jun 17 '13 at 22:25
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    $\begingroup$ The containment "continuous"$\subset$"integrable" depends on the domain of integration: It is true if the domain is closed and bounded (a closed interval), false for open intervals, and for unbounded intervals. $\endgroup$ – Andrés E. Caicedo Jun 17 '13 at 22:51
  • $\begingroup$ Yes.. Thats an important point to keep in mind... got it all now...thanks yet once again :)) $\endgroup$ – under root Jun 17 '13 at 23:02
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Let $g(0)=1$ and $g(x)=0$ for all $x\ne 0$. It is straightforward from the definition of the Riemann integral to prove that $g$ is integrable over any interval, however, $g$ is clearly not continuous.

The conditions of continuity and integrability are very different in flavour. Continuity is something that is extremely sensitive to local and small changes. It's enough to change the value of a continuous function at just one point and it is no longer continuous. Integrability on the other hand is a very robust property. If you make finitely many changes to a function that was integrable, then the new function is still integrable and has the same integral. That is why it is very easy to construct integrable functions that are not continuous.

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    $\begingroup$ continuity is fragile, integrability is robust. what's somethin antifragile? $\endgroup$ – BCLC Sep 28 '16 at 7:16
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Probably the simplest example of an integrable function that's not continuous is something like $$f(x)=\left\{\begin{array}{rl}3 & 0\leq x<1\\ 5 & 1\leq x\leq 2.\end{array}\right.$$

This $f$ is clearly not continuous at 1, but it is Riemann integrable on $[0,2]$, with $\int_0^2 f(x)\ dx = 8$.

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In order for some function f(x) to be continuous at x = c, then the following two conditions must be true: f(c) is defined and the limit of f(x) as x approaches c is equal to f(c). Recall that the limit of a polynomial p(x) as x approaches c is p(c), therefore polynomials are always continuous.

In order for some function f(x) to be differentiable at x = c, then it must be continuous at x = c and it must not be a corner point (i.e., it's right-side and left-side derivatives must be equal).

Continuity implies integrability; if some function f(x) is continuous on some interval [a,b], then the definite integral from a to b exists. While all continuous functions are integrable, not all integrable functions are continuous. This is because the limit L of some function f(x) as x approaches c can exist despite discontinuity at x = c, so long as f(x) approaches L from both sides of x. Recall that the definition of the definite integral is based upon a Riemann Sum as n tends to infinity (or dx tends to zero, if you prefer) using the limit process. Therefore, any behavior of the function that would cause the limit to not exist would also cause the function to not be integrable. Recall that there are three situations that commonly cause a limit to not exist: different right-side and left-side limits, oscillation, and asymptotes or unbounded/infinite curves.

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$[x]$ (the integer part of $x$) is integrable over $[0, 3]$, but it is not continuous at $1$ and $2$, therefore it is not continuous on $[0,3]$.

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