Problem 1: $\Omega$ open, $\varphi:[0,1]\to\Omega$ is a closed path, define $\varphi_n:[0,1]\to\Omega$ to be linear on $\left[\dfrac{j-1}{n},\dfrac{j}{n}\right]$, $\varphi_n\left(\dfrac{j}{n}\right)=\varphi\left(\dfrac{j}{n}\right)$. Show $\exists$ $n_0$ such that for $n>n_0$, $\varphi_n([0,1])\subseteq \Omega$ and $\varphi_n \cong \varphi$ (i.e. $\cong$ is notation for homotopy).
Proof of $\varphi\cong \varphi_n$: Since $\varphi$ is a closed path, $\varphi(0)=\varphi(1)$. We construct \begin{gather*} H: [0,1]\times [0,1]\to\Omega \\ H(s,t)=s\varphi(t)+(1-s)\varphi_n(t) \end{gather*} where $s,t\in [0,1]$. We need to verify the conditions of a homotopy
- $H(0,t)=0\cdot\varphi(t)+(1-0)\varphi_n(t)=\varphi_n(t)$
- $H(1,t)=1\cdot\varphi(t)+(1-1)\varphi_n(t)=\varphi(t)$
- $H(s,0)=s\varphi(0)+(1-s)\varphi_n(0)=s\varphi(0)+(1-s)\varphi(0)=s\varphi(1)+(1-s)\varphi(1)=s\varphi(1)+(1-s)\varphi_n(1)=H(s,1)$
Hence $\varphi\cong\varphi_n$. Is this proof correct?
Also, how would I show that $\varphi_n([0,1])\subseteq\Omega$?
Problem 2: If $\varphi$ rectifiable, $f\in\mathcal{O}(\Omega)$, then $\int\limits_{\varphi_n} f \to \int\limits_{\varphi} f$. (independent of $n$)
Doesn't this follow immediately from Problem 1?
You are missing some important points. You have not even taken n large enough. You have to show that $\phi_n$ has range in $\Omega$ but more importantly you have to make sure that $H$ also has its range in $\Omega$. Only then you can say that $\phi_n$ and $\phi$ are homotopic in $\Omega$. Do prove that let $\delta =d(\phi ([0,1]),\mathbb C - \Omega)$. this is a positive number. Now that fact that $\phi$ is uniformly continuous implies that $\phi_n \to \phi$ uniformly. If $|\phi_n (t) \phi (t)| < \delta$ it is easy to see that $\phi_n ([0,1])$ is contained in $\Omega$. A similar argument shows that H also takes values in $\Omega$. (I will give more details if you ask).