I have formulated the following proof for proving completeness of a $L_{\infty}$ space. Kindly verify the same:
Given $f_n$ is a cauchy sequence in $L_\infty$. Prove that it converges to an element of $L_\infty$.
$<=> \forall\varepsilon>0 \ (\exists N_\varepsilon \ \epsilon \ \Bbb N \ (\forall (m,n) \ (min(m,n)> N_\varepsilon\Rightarrow ||f_m - f_n||_\infty < \varepsilon)))$
Define $$B^{(m,n)}_\varepsilon = \{x: \ |f_m(x)-f_n(x)|\geq\varepsilon\}$$ $$B_\varepsilon=\cup_{min(m,n)>N_\varepsilon}B^{(m,n)}_\varepsilon$$
Then $min(m,n)>N_\varepsilon \Rightarrow (||f_m-f_n||_\infty<\varepsilon) \Rightarrow (\mu(B^{(m,n)}_\varepsilon)=0)$
$$\therefore \mu(B_\varepsilon)=0$$
Let $x \epsilon B^{c}_\varepsilon$ where the c in superscript denotes set compliment.
Then $min(m,n)> N_\varepsilon\Rightarrow |f_m(x)-f_n(x)|<\varepsilon$
Put $B = \cup_{k\epsilon \Bbb N}B_{1/k}$ then $\mu(B)=0$
Then $x \epsilon B^c$ and $\varepsilon>0$ $$\exists k \epsilon \Bbb N, \varepsilon> 1/k$$ $$min(m,n)>N_{1/k} \Rightarrow |f_m(x)-f_n(x)|< 1/k < \varepsilon$$ $\therefore$ the sequence $f_n$ converges uniformly on $B^c$.
Let $f_n(x) \rightarrow f(x)$ for $x \epsilon B^c$. Define $f$ to be $0$ on $B$.
Now for $\forall x \epsilon B^c(\forall \varepsilon>0, \exists N_{\varepsilon} \epsilon \Bbb N(min(m,n)>N_{\varepsilon} \Rightarrow |f_m(x)-f_n(x)|< \varepsilon))$
Letting $n \rightarrow \infty$ we get $\forall m >N_{\varepsilon}, |f_m(x)-f(x)| \leq \varepsilon, x \epsilon B^c$ i.e. $||f_m(x)-f(x)||_{\infty} \leq \varepsilon$.
Also $||f||_\infty < |f_m(x)|+\varepsilon$, $\therefore f \epsilon L_\infty$
Hence we have shown that if $f_n$ is cauchy in $L_\infty$ then it converges to an element of $L_\infty$ in $L_\infty$ norm.