Show that $L^{\infty}$ is a Banach Space with respect to the norm $||.||_{\infty}$ where $||f||_{\infty}=\inf\{ a\ge 0: \mu\left(\{x: |f(x)| \gt a\}\right)=0\}$
I am going to prove the following lemmas and then use them in the proof.
Lemma-1: $||.||_{\infty}$ is a norm on $L^{\infty}$
Proof: Suppose that $||f||_{\infty}=0$. Let $A=\{x \in X : f(x) \ne 0\} $ and $A_n=\{x \in X: |f(x)| \gt \frac{1}{n}\}$ . Then $A=\cup_{n}A_n$. Then there exists $'a'$ such that $\mu\left(\{x: |f(x)| \gt a\}\right)=0$ and $a \lt \frac{1}{n}$ (by the definition of $||f||_{\infty}$). Thus $$\mu\left(\{x: |f(x)| \gt \frac{1}{n}\}\right) \le \mu\left(\{x: |f(x)| \gt a\}\right)=0$$
$\implies \mu(A_n)=0$ and hence $\mu(A)=0$. Thus $f =0 ,\mu$-a.e.
Now $$|f(x)+g(x)| \le |f(x)| +|g(x)| \le ||f||_{\infty}|+||g||_{\infty} ,\mu-\text{a.e}$$
Thus $||f+g||_{\infty} \le ||f||_{\infty}+||g||_{\infty}$
Now $$||\lambda f||_{\infty}=\inf\{a \ge 0: \mu\left(\{x: |\lambda f(x)| \gt a\}\right)=0\}$$ $$=\inf\{|\lambda|a \ge 0:\mu\left(\{x: |\lambda||f(x)| \gt |\lambda|a\}\right)=0\}$$ $$=|\lambda|\inf\{a \ge 0:\mu\left(\{x: |f(x)| \gt a\}\right)=0\}=|\lambda|||f||_{\infty}$$
Lemma-2: $||f_n-f||_{\infty} \to 0$ iff there exists $E \in \mathcal{M}$ such that $\mu(E^c)=0$ and $f_n \to f$ uniformly on $E$.
Proof: $(\implies)$ For every $k \in \mathbb{N}$, there exists $n_0(k) \in \mathbb{N}$ such that for all $n \ge n_0(k)$, $||f_n-f||_{\infty} \lt \dfrac{1}{k}$. Then for each $n \ge n_0(k)$ there exists $E_n^{k}$ such that $\mu(E_n^k)=0$ and $|f_n(x)-f(x)| \lt \dfrac{1}{k}$ for all $x \in (E_n^k)^{c}$. Let $$E^k=\cup_{n \ge n_0(k)} E_n^k$$. Then $\mu(E^k)=0$ and for all $x \not \in E^k, |f_n(x)-f(x)| \lt \frac{1}{k}, \forall n \ge n_0(k)$. Now Let $E=\cup_{k=1}^{\infty} E^k$. Then $\mu(E)=0$. Let $\epsilon \gt 0$. Then there exists a $k_0 \in \mathbb{N}$ such that $\frac{1}{k} \lt \epsilon$ for all $k \ge k_0$. Let $x \in E^c$. In Particular $x \not \in E^{k_0}$ . Then for all $n \ge n_0(k_0), |f_n(x)-f(x)| \lt \frac{1}{k_0} \lt \epsilon$
($\impliedby$) Let $\epsilon \gt 0$. Then there exists $n_0 \in \mathbb{N}$ such that for all $n \ge n_0, |f_n(x)-f(x)| \lt \epsilon, x\in E$. Then for all $n \ge n_0$,
$$\left(\{x: |f_n(x)-f(x)| \gt \epsilon\}\right) \subset E^c$$ $$\implies \mu\left(\{x: |f_n(x)-f(x)| \gt \epsilon\}\right)=0 $$ which in turn gives us that for all $n \ge n_0$, we have $$||f_n-f||_{\infty} \lt \epsilon$$.
(Proof that $L^{\infty}$ is a Banach Space) :
Let $\{f_n\}_{n \in \mathbb{N}} \in L^{\infty}$ be a cauchy sequence. Then for $\epsilon \gt 0$, there exists a $n_0(\epsilon) \in \mathbb{N}$ such that for all $n,m \ge n_0(\epsilon)$, we have $||f_n-f_m||_{\infty} \lt \dfrac{\epsilon}{2}$. Then there exists $E^{\epsilon} \in \mathcal{M}$ such that $\mu(E^{\epsilon})=0$ and for $x \in (E^{\epsilon})^{c} ,|f_n(x)-f_m(x)| \lt \dfrac{\epsilon}{2}$.
Let $$E= \cup_n E^{\frac{1}{n}}.$$ Then $\mu(E)=0$ and for all $x \in E^c, \{f_n(x)\}$ is a cauchy sequence. Let $f(x)=\lim_n f_n(x)$ for $x \in E^c$. Let $\epsilon \gt 0$. Then there exists $k_0 \in \mathbb{N}$ such that for all $k \ge k_0,\frac{1}{k} \lt \frac{\epsilon}{2}$. Then for all $n, m \ge n_0(k_0)$, we have $$|f_n(x)-f_m(x)| \lt \frac{1}{k} \lt \frac{\epsilon}{2}$$ Letting $m \to \infty$ we have $$|f_n(x)-f(x)| \lt \epsilon, x\in E^c.$$ By Lemma-2, we have $||f_n-f||_{\infty} \to 0$. From here by use of triangle inequality, we see that $ f\in L^{\infty}$.
Is this proof alright?
Thanks for the help!!