1. ## Re: Discontinuity and Integral

But you didn't show that the function is defined for irrational numbers, and you didn't use the standard definition of an integral as a limit of partitions, which I thought was the subject of this thread, so the example is meaningless.

But you win. I am now replying from the local library since after I switched from MHF back to the internet from my last reply to Plato I got a blank screen and now can't get back on. Restarting my modem usually works, but now doesn't. Looks like I'm in for massive inconvenience. Is that your answer to my questions?

Having my computer infected is better than being banned because there is no onus attached to it.

Hope I haven't shut down the local library.

2. ## Re: Discontinuity and Integral

Originally Posted by Hartlw
But you didn't show that the function is defined for irrational numbers, and you didn't use the standard definition of an integral as a limit of partitions, which I thought was the subject of this thread, so the example is meaningless.

But you win. I am now replying from the local library since after I switched from MHF back to the internet from my last reply to Plato I got a blank screen and now can't get back on. Restarting my modem usually works, but now doesn't. Looks like I'm in for massive inconvenience. Is that your answer to my questions?

Having my computer infected is better than being banned because there is no onus attached to it.

Hope I haven't shut down the local library.
take another look at the function definition. The function isn't even discontinuous at the irrationals. It's discontinuous at the set

$\{x : x = \dfrac 1 {n+1} ~~n \in \mathbb{N}\}$

$\dfrac 1 {n+1}\in \mathbb{Q}$

3. ## Re: Discontinuity and Integral

Finally got home computer accessing the Internet again- quite a scare. But not taking any chances and using a public computer to access MHF.

From Shilov, “Elementary Real and Complex Analysis,” Dover, 9.17

1) “..every bounded function with only finitely many points of discontinuity is integrable.”

2) “f(x) has a Riemann integral iff the set of all its points of discontinuity can be covered by finite or countable set of intervals whose total length is δ, where δ is arbitrary.”

The only kind of discontinuity I can imagine for 2) is x=1 for x rational and 0 for x irrational in the interval limit -1/n < x0 < 1/n, but that doesn’t work because the only number in all the intervals is x0.

Finally, other respondents may be talking about Lebesque integration without saying so, but I can’t imagine responsible moderators on MHF being that devious under the guise of being “advanced.”

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