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Formal definition of the integral?

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I believe there are also such functions with infinity amount of points(Countable, and in a finity fragment) with no continuity, which you can calculate their rieman integral.
The definition of a Riemann Integral requires a function to be defined at all points on an interval.


Sigh. It is the case that Riemann-integrable functions may be discontinuous at an infinite number of points. A correct statement is: A Riemann-integrable function is continuous except possibly on a set of measure zero.

Intelligent discussion of the limits of Riemann integration requires an understanding of the Lesbeque and the Stieltjes integrals (they "solve" different "problems" with Riemann's version).

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Original post by sherifffruitfly
False. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.

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Original post by NotAnAnonymousPoster
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Original post by sherifffruitfly
False. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.


You can say that all you want - it's just false. Definitions do not vary on this to any substantial degree (and hence neither to the theorems). ROFL - you think there's substantial variance in exactly *what* the Riemann integral is? LMAO.

And there's nothing "degenerate" about my example. It's a perfectly fine continuous, differentiable, and Riemann-integrable function.

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Original post by sherifffruitfly
You can say that all you want - it's just false. Definitions do not vary on this to any substantial degree (and hence neither to the theorems). ROFL - you think there's substantial variance in exactly *what* the Riemann integral is? LMAO.
Not for the Riemann integral, no. Which is why I said:
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Original post by Me
I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.
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Original post by sherifffruitfly
And there's nothing "degenerate" about my example. It's a perfectly fine continuous, differentiable, and Riemann-integrable function.
I wasn't clear there. I meant that your function f is defined on the degenerate interval [0,0].

But we won't get anywhere here without actually supplying some definitions. Try this one at PlanetMath:

Riemann Integral

The definition requires that the function in question be defined on a non-degenerate interval.

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Original post by NotAnAnonymousPoster
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Original post by sherifffruitfly
False. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.


Moreover, the theorem I was thinking of was not the one you tried to putin my mouth. The one I was after was the one I actually said - that the Riemann-integrable functions are those which are continuous except on a set of measure zero.

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But we won't get anywhere here without actually supplying some definitions. Try this one at PlanetMath:

Riemann Integral

The definition requires that the function in question be defined on a non-degenerate interval.


Kindly point out *exactly*, by quoting, where in the PM definition it is "required" (required for what?) for an interval (what interval?) to be "non-degenerate" (what does non-degenerate mean?).

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Original post by sherifffruitfly
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But we won't get anywhere here without actually supplying some definitions. Try this one at PlanetMath:

Riemann Integral

The definition requires that the function in question be defined on a non-degenerate interval.


Kindly point out *exactly*, by quoting, where in the PM definition it is "required" (required for what?) for an interval (what interval?) to be "non-degenerate" (what does non-degenerate mean?).
Looking again, it probably doesn't require it. A degenerate interval is an interval [a,a]={a}. A non-degenerate interval is not a degenerate interval. But this is moot anyway.

The PM definition defines Riemann-integration on intervals. You said that your function f is not defined on an interval, so its nowhere Riemann integrable by the PM definition.

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The PM definition defines Riemann-integration on intervals. You said that your function f is not defined on an interval, so its nowhere Riemann integrable by the PM definition.


Apparently you're under the massive misconception that [a,b] is not an interval when a=b. Well, it is. It is a closed, bounded, compact, finite, perfect interval. I'm sure there are a number of other qualities such intervals have as well.

Moreover, it pretty much *has* to be this way - if [a,a] weren't an interval, then all sorts of topological and lattice-theoretic things would f^ck up (closure under arbitrary intersections being the most obvious).

There's absolutely nothing degenerate about my example, and my theorem was correct as I originally stated: f Riemann-integrable = f continuous but at a set of measure 0.

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ervals. You said that your function f is not defined on an interval, so its nowhere Riemann integrable by the PM definition.


rofl - and although it must be blasphemous to someone who thinks as you do, the empty set is also an interval.

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Original post by sherifffruitfly
Apparently you're under the massive misconception that [a,b] is not an interval when a=b. Well, it is. It is a closed, bounded, compact, finite, perfect interval. I'm sure there are a number of other qualities such intervals have as well.

Moreover, it pretty much *has* to be this way - if [a,a] weren't an interval, then all sorts of topological and lattice-theoretic things would f^ck up (closure under arbitrary intersections being the most obvious).

There's absolutely nothing degenerate about my example, and my theorem was correct as I originally stated: f Riemann-integrable = f continuous but at a set of measure 0.
Great. Now can you retract your original objection, and accept that your new position is in contradiction with it:
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Original post by sherifffruitfly
A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
Your function's domain is and includes the interval [0,0].

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