functional analysis – Boundedness of the multiplication operator

Let $(Omega,mathcal{F},mu)$ be a measure space and $f: Omega to mathbb{C}$ be a measurable function. Let $M_f$ be the multiplication operator whose domain is those $gin L^2(Omega,mathcal{F},mu)$ for which $fg in L^2(Omega,mathcal{F},mu).$

If $f$ is essentially bounded, then I can prove that $M_f$ is a bounded operator with $|M_f|=|f|_{infty}$ where $|f|_{infty}$ is the essential supremum of $f.$

I am unable to prove the converse! Here is my attempt.

Suppose $f$ is not essentially bounded, then $E_n={w inOmega: |f(w)| geq n}$ has positive measure for all $n in mathbb{N}.$ So, if I can find for every $n$ a non-zero $g$ in the domain of $M_f$ that vanishes outside $E_n$ then we get
|M_f(g)|&= int_{Omega} |fg|^2 \
&= int_{E_n} |f|^2 |g|^2 \
&geq n^2 |g|^2

which implies $|M_f| geq n$ proving that $M_f$ is not bounded.

But I don’t see how I can find such $g.$

P.S. If we assume that $(Omega,mathcal{F},mu)$ is $sigma-$finite, then I can find a subset of $F_{n,m}$ of $E_n$ of finite positive measure with $n leq |f|< n+m$ on $F_{n,m},$ and let $g=mathbb{1}_{F_{n,m}}.$