# If a function is continuous everywhere, but undefined at one point, is it still continuous?

This is a question regarding the definition of continuity.

My understanding of continuity is that a function is continuous at a point when it holds that $$\lim_{x\to a^-}f(x) = f(a) = \lim_{x\to a^+}f(x) \quad \quad (1)$$

The book I'm currently reading has this image: Note here that $$f(x)$$ is defined for $$x=3$$, but $$g(x)$$ is not.

This is followed by text stating that

g(x) is continuous because $$D_g = [0, 6]\text{\\}\{3\}$$, thus it is continuous for all values in its domain.

My point of contention here is that, how can we say that it is continuous at $$x=3$$ when $$g(3)$$ does not exist? Referring to the aforementioned definition $$(1)$$ that the limits converge to the actual value at this point.

I would have immediately declared both cases as jump discontinuities.

Am I mistaken here? Does $$g(x)$$ illustrate an exception to $$(1)$$?

• It is nowhere said that $g$ is continuous at $x=3$. – Martin R Nov 20 '18 at 10:01
• There is a theorem saying that $g:D\to\cdots$ is continuous iff $g$ is continuous at every $x\in D$. Here $3\notin D$ so is irrelevant if it comes to the question whether $g$ is continuous. See here for a related question. – drhab Nov 20 '18 at 10:06
• Also potentially relevant: If there are finitely many points where there are discontinuities, then $g$ is continuous almost everywhere. – Spitemaster Nov 20 '18 at 19:06
• If a function is continuous everywhere, but undefined at one point, the sky is purple. – user253751 Nov 20 '18 at 21:44

$$G$$ is continuous on the domain $$[0,3)\cup(3,6]$$.
3 is not in the domain. For every point in the domain of $$g$$, we have the required convergence.
The function is continuous everywhere in the interval except that point deleted from the domain, it's more a nuance of the language than anything else. Choose any point that is not $$3$$ in that interval: you can then find left- and right-hand limits to that point and show they're equal.