I have a question regarding a proof regarding lower semicontinuous functions (the proof of the claim below).
Definition: We call a function $f\colon \mathbb{R}^n\to\mathbb{R}\cup \{\infty\}$ lower semicontinuous if for each $k\in\mathbb{R}$ the set $\{x\in\mathbb{R}^n\mid f(x)>k\}=:M_k(f)$ is open in $\mathbb{R}^n$.
Claim: If $(f_n)_n$ is a sequence of lower semicontinuous functions such that $f_n\uparrow f$ pointwise. Then $f$ is lower semicontinuous.
Proof: Let $z\in M_k(f)$. We proof that there exists $\epsilon >0$ such that $B_\epsilon (z)\subset M_k(f)$.
Set $\delta:=f(z)-k$. Then $\delta >0$. Since $f_n(z)\uparrow f(z)$ and $f(z)>k$, there exists $n_0\in \mathbb{N}$ such that $f_{n_0}(z)-\frac{\delta}{2}>k$. Since $f_{n_0}$ is lower semicontinuous, there exists $0<\epsilon <\frac{\delta}{2}$ such that $B_\epsilon(z)\subset M_k(f_{n_0})$. Let $y\in B_\epsilon(z)$. Then it is $f(y)=f_{n_0}(y)>k$, since $\epsilon <\frac{\delta}{2}$. Thus, $B_\epsilon(z)\subset M_k(f)$, i.e.f is lower semicontinuous.
My questions:
1) where do we need the explicit definition of $\delta$ i.e. $\delta=f(z)-k$?
2) Applying that $f_{n_0}$ is lower semicontinuous, how to we know that we find that $\epsilon$ is between $0$ and $\frac{\delta}{2}$?
3)I don't unserstand why it is $f(y)=f_{n_0}(y)$ for this $y$.
This proof is a student solution from an exercise class a year ago (so that maybe there are some mistakes). I'm not sure if I'm too stupid or if there are mistakes. If these are mistakes, how can one fix it, or how else can be proven that there exists $\epsilon >0$ such that $B_\epsilon (z)\subset M_k(f)$?
The proof seems garbled. Here is how I would do it: Let $\epsilon>0$ and $x_0\in \mathbb R^n$. Choose an integer $N$ such that $f(x_0)-f_N(x_0)<\epsilon/2.$ For this $f_N$, choose $\delta>0$ such that $f_N(x)>f_N(x_0)-\epsilon/2$ whenever $|x-x_0|<\delta.$ Then, for this $\delta,$ $f(x)>f_N(x)>f_N(x_0)-\epsilon/2>f(x_0)-\epsilon/2-\epsilon/2=f(x_0)-\epsilon. $