Continuity at a point implies integrability around point?

In summary, the conversation discussed the question of whether a function that is continuous at a point must also be integrable on a closed interval containing that point. The definition of integrability provided by Spivak was used, and a possible proof was presented, but it was found to contain errors. The conversation then shifted to discussing examples of nonintegrable functions and the need for a fixed interval for proving integrability.
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If a function f is continuous at a point p, must there be some closed interval [a,b] including p such that f is integrable on the [a,b]?

As a definition of integrable I'm using the one provided by Spivak: f is integrable on [a,b] if and only if for every e>0 there is a partition P of [a,b] such that U(f,P)-L(f,P)<e, where U denotes an upper sum and L denotes a lower sum.

Here is what I think is a proof, but which probably contains some error:

Since f is continuous at a point p, there is some s'>0 such that for every point x, if |x-p|< s' then | f(x) - f(p) | < e/2 for arbitrary e>0. Denote s=min(s', 1). Choose points a and b such that (p-s) < a < p < b < (p+s) Let a be a point with a<p and and p-a<s. Let b be a point with b>p and b-p<s.

For our partition P, we can use P={a,b}.
Clearly, L(f,P) > s ( f(p) - e/2 ) and U(f,P) < s ( f(p) + e/2 ).
So U(f,P)-L(f,P) < s*e < e.

Thus f is integrable on [a,b].

If this proof works, then it provides an easy way of proving that continuity implies integrability on an interval [a,b]. You just examine
z = sup {x: a≤x≤b and f is continuous on [a,x] }. Since f is also continuous at z, there is some interval around z [p,q] which is integrable. Since [a,p] is integrable and so is [p,q], [a,q] is integrable, which contradicts the fact that z is the least upper bound. Thus z=b.
 
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  • #2
It doesn't work. To prove integrability on an interval you have to fix the interval. Then show for any e>0 you can find such a partition. The interval can't depend on e. I would look for a counterexample. What kind of examples of nonintegrable functions do you know?
 

1. What does the concept of continuity at a point mean?

The concept of continuity at a point refers to the idea that a function is smooth and has no abrupt changes or breaks at a specific point. This means that as the input of the function approaches the point, the output also approaches a specific value, without any sudden jumps.

2. How is continuity at a point related to integrability around a point?

Continuity at a point is a necessary condition for a function to be integrable around that point. This means that if a function is continuous at a point, it can be integrated around that point to find the area under the curve.

3. What is the significance of continuity at a point in calculus?

In calculus, continuity at a point is important because it allows us to apply fundamental concepts such as limits, derivatives, and integrals. It also helps us understand the behavior of a function at a specific point, which is crucial in many real-world applications.

4. Can a function be continuous at a point but not integrable around that point?

No, a function cannot be continuous at a point but not integrable around that point. This is because a continuous function is defined as one that can be drawn without lifting the pencil from the paper, and such a function can always be integrated to find the area under the curve.

5. Are there any other conditions besides continuity at a point for a function to be integrable around that point?

Yes, besides continuity, a function must also be bounded around the point and have a finite number of discontinuities in order to be integrable around that point. This means that the function cannot have any infinite or undefined values, and cannot have an infinite number of breaks or jumps at the point.

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