Relationship between Simply Connectd Domains, Cauchy's Theorem, and Jordan curves For a while, before seeing the formal statement of Cauchy's theorem, I used to think that the following is actually Cauchy's Theorem:

Let $C$ be a rectifiable (or piecewise smooth) Jordan curve $C$ and let $\Omega$ be a domain containing $C$ and its interior. If $f$ is holomorphic on $\Omega,$ then $\int_C f=0.$

Now, of course I know that the requirenment is actually for $\Omega$ do be simply connected. But intuitively, it does seem like the interior of a Jordan curve is probably connected, and that we may extend the interior a bit to cover the curve itself, and still have a simply connected domain.
So I am wondering, is my old version of Cauchy's Theorem actually true?
 A: You say you know that the requirement for $\int_C f = 0$ is that $\Omega$ is simply connected. In fact, if $\Omega$ is simply connected, then for all holomorphic $f : \Omega \to \mathbb C$ and all closed curves $C$ in $\Omega$ we have $\int_C f = 0$. However, this is just a special case of the following:
If $C$ is a closed curve in $\Omega$ which is null-homotopic in $\Omega$, then $\int_C f = 0$ for all holomorphic $f : \Omega \to \mathbb C$.
Now any Jordan curve in $\Omega$ such that $\Omega$ contains the interior of $C$ is null-homotopic in $\Omega$.
In fact, this is a consequence of the Schoenflies Theorem (see e.g. https://en.wikipedia.org/wiki/Schoenflies_problem). This says that there exists a homeomorphism $h : \mathbb C  \to \mathbb C$ such that $h(S^1) = C$ (here $S^1$ is the standard circle). Let $B$ denote the open unit disk in $\mathbb C$, $B' = \mathbb C \setminus (B \cup S^1)$, $U$ the interior of $C$ and $U' =  \mathbb C \setminus (U \cup C)$. Then $B, B'$ are the components of $\mathbb C \setminus S^1$ and $U, U'$ are the components of $\mathbb C \setminus C$. Hence $h$ maps $B$ either onto $U$ or onto $U'$. But $D = h(B \cup S^1) = h(B) \cup C$ is compact which is possible only when $h(B) = U$. So $D = U \cup C$ is contained in $\Omega$ and, being homeomorphic to the closed unit disk $D^2 = B \cup S^1$, it is contractible. Hence $C$ is null-homotopic in $D \subset \Omega$.
A: Here's an interesting, different approach from that of Paul Frost.
There is a slightly stronger version of Cauchy's Theorem (the homological one) which states the following:
Theorem:
Let $\Omega\subset\mathbb{C}$ be an open set and $f$ a holomorphic function in $\Omega$. If $C$ is a rectifiable closed curve that satisfies
$$\operatorname{Ind}(C;a)=0 \text{ for all }a\in\mathbb{C}\setminus\Omega$$
then
$$\int_C f = 0$$
In your case, $C$ is a Jordan curve, so it splits the complex plane in two connected components: the interior (which is contained in $\Omega$) and the exterior (which therefore contains $\mathbb{C}\setminus\Omega$). It is easy to see that the winding number ($\operatorname{Ind}(C;\cdot):\mathbb{C}\setminus C\to\mathbb{Z}$) is continuous, hence locally constant, in particular it is constant in each connected component of $\mathbb{C}\setminus C$. It is also easy to see that if $D$ is a disk that contains $C$, $\operatorname{Ind}(C;a)=0$ $\forall\ a\in\mathbb{C}\setminus D$. Putting it all together, we get that $\operatorname{Ind}(C;a)=0$ for all $a$ in the exterior of $C$ (which contains $\mathbb{C}\setminus\Omega$). Using the above version of Cauchy's Theorem, we easily get that $\int_C f = 0$.
