# Function holomorphic except for real line and continuous everywhere is entire

1) I've already shown that if $f:\mathbb{C}\rightarrow\mathbb{C}$ is holomorphic everywhere except for a single point and if it continuous on whole $\mathbb{C}$ then it is entire. It was quite easy.

But now I have to generalize it onto a whole real axis in $\mathbb{C}$. Can you help me with that?

2) On the other hand I wonder if it can be generalized even more, for even bigger sets?

I mean what are the "biggest" sets $S$ that apply to the following statement?

If $f:\mathbb{C}\rightarrow\mathbb{C}$ is holomorphic on $\mathbb{C} \setminus S$ and continuous on $\mathbb{C}$ then it is holomorphic everywhere on $\mathbb{C}$.

Use Cauchy's theorem combined with Morera's theorem. The integrals along any triangle contained in one of the open semiplanes is zero due to Cauchy. It remains to prove that the integrals along triangles that intersect the line are zero too.

Decomposing triangles into smaller triangles you can reduce to the case in which the triangle has an edge on the line. Now decompose this triangle into many many little triangles. All triangles that don't touch the line are zero. The ones that do touch are very little, so by continuity the integral on them is arbitrarily small. Take limit as the number of triangles growth and you are done.

[In the image imagine the big triangle to have constant size.] The characterization of the sets that you can remove is more difficult. There is no good geometric characterization. But they can be characterized using the concept of analytic capacity (See the last section).

• Yes, I get the idea. But writing it formally can take a while... – luka5z Jan 13 '15 at 16:44
• @luka5z There is no much difference between this and the formal proof. Just give names to the circuits and claim, by Cauchy, that those that don't intersect the line have integral zero. – Pp.. Jan 13 '15 at 16:46
• Ok, I get it now, thanks! – luka5z Jan 13 '15 at 16:49
• This is related. – Pedro Tamaroff Jan 13 '15 at 16:54