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Can the canonical theorem on Riemann integrability (namely, that a bounded real-valued function defined on a closed interval is Riemann-integrable if, and only if, its discontinuities constitute a set of measure zero) be taken as the definition of continuity? If so, it would then be a theorem that a function f is continuous at a limit point a in its domain if, and only if, the limit of f(x) as x goes to a is f(a).

edit (2.Sep.2013, CST, MERCA)

I am not interested in defining continuity this way. I'm only interested in whether it is POSSIBLE to define continuity this way. That is, just as there is a (long) list of statements equivalent to the Axiom of Choice, so also was I wondering whether this is on the list of statements equivalent to the definition of continuity, that's all. I (weakly) conjecture that it is indeed on the list, and so I'm holding out for an answer that proves this. An answer that simply says that the answerer does not see how it can be done is not sufficient for me to give up my conjecture, but a counterexample certainly would.

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I think you're going to have to clarify how you expect a definition of continuity to arise out of this. –  Ben Millwood Sep 2 '13 at 0:15
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To elaborate on Ben's point: this theorem lets us make the following definition: A bounded function $f\colon [a,b] \to \mathbb{R}$ is "continuous almost everywhere" iff it is Riemann integrable. However, it's not clear (at least to me) how to proceed to define the notion of continuity at a point. (And on top of that, there's the fact that the theorem is limited to bounded functions.) The underlying issue, I think, is that given a function $f\colon [a,b] \to \mathbb{R}$ altering its value at a single point will affect its continuity but not its Riemann integrability –  Jesse Madnick Sep 2 '13 at 2:07
    
@Mike: The Axiom of Choice is a statement -- it has a truth value -- as are the things that are equivalent to it. What does it mean for two definitions (of different things) to be equivalent (definitions can't be true or false)? Can you give an example of this? Until you do this what you are asking does not seem clear enough to allow a proof or counterexample. –  Pete L. Clark Sep 2 '13 at 16:04
    
To say a little more: sometimes the definition of $X$ can be expressed in terms of the definition of $Y$ and conversely, in a useful way. For instance, the definition of continuity is traditionally given in terms of limits. But this can also be done the other way around: $\lim_{x \rightarrow c}f(x) = L$ if when you define $f(c) = L$ you get a function which is continuous at $c$. This notion of equivalence is useful but very informal. I'm not sure how meaningful it is to conjecture such an equivalence exists because I don't know what it means to disprove such a conjecture.... –  Pete L. Clark Sep 2 '13 at 16:15
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...For instance, suppose I said "I conjecture that continuity of a function $f: \mathbb{R} \rightarrow \mathbb{R}$ can be given in terms of cyclic groups". That sounds ridiculous, but how do I disprove it? –  Pete L. Clark Sep 2 '13 at 16:16

2 Answers 2

The (Riemann)-Lebesgue Criterion gives a criterion for a bounded function to be almost everywhere continuous. So you could take that as the definition of almost everywhere continuity for bounded functions if you want, although I don't see any particular advantage in doing so. (Why do you want to define continuity in terms of Riemann integrability anyway?)

If $f$ is continuous at $c$ and we modify the value of $f$ at $c$ (while keeping every other value unchanged) then the modified function cannot be continuous at $c$. The same holds if we modify the value of $f$ at any finite set of points on its domain.

On the other hand, if $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable, then if we change the values of $f$ at any finite set we do not disturb the Riemann integrability or the value of the integral. This is a nice elementary exercise involving the definition of the Riemann integral: the idea is to take an arbitrarily small subinterval around each modified point.

These diametrically opposite properties make it hard to see how one could possibly define continuity in terms of Riemann integrability.

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This answer is mostly pretty good, but I find your description of continuity as a pointwise property a rather strange one. After all, you can't tell if a function is continuous by looking at any single point (heck, in the real case you can't even look at merely countably many points!) – it's only when you look at relationships between points that you can see it. Contrast to a property like "pointwise bounded" where the property holds exactly if it holds at every point considered individually. –  Ben Millwood Sep 2 '13 at 5:08
    
@Ben: OK, I agree that "pointwise property" didn't really mean much, so I took it out and left the contrasting properties of continuity and Riemann integrability that I actually meant. –  Pete L. Clark Sep 2 '13 at 5:12

You have given a definition of Riemann integrability for bounded functions, which states that the function f can be discontinuous on a set of measure 0. So if it is discontinuous, it is discontinuous, even though it is perfectly integrable. Your definition of continuity is correct, but there is no guarantee that f will meet that definition -- it might, or it might not.

However $\int f$ (i.e. any function F whose derivative is f where f is continuous) can be defined to be continuous although it will not be everywhere differentiable.

In general integration makes functions smoother, so that a slightly discontinuous function has a continuous integral F (in the sense of described above); a continuous function will yield an F that is differentiable and so on.

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