If the multiplication operator has full domain, the multiplier is essentially bounded.

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From: Norm of multiplication operator in $L_p$ [NOTE: Unlike this question, here, $p$ can be $\infty$.]

Let $Ω$ be an open set in $R^n$, and let $a$ be a measurable complex-valued function on $Ω$. The (maximal) operator of multiplication by $a$ on $L_p(Ω) \ (1 ≤ p ≤ ∞)$ is defined by

$M_au = au, u ∈ dom(M_a)$

$dom(M_a) = \{u ∈ Lp(Ω):\ au ∈ L_p(Ω)\}.$

The multiplication operator $M_a$ defined on the whole space $L_p(Ω)$ if and only if $a$ is essential bounded, that is, $a ∈ L_∞(Ω)$.

I have to show that $dom(M_a)=L_p(Ω) \Rightarrow a ∈ L_∞(Ω)$, and I found the above question. I read MULTIPLICATION AND COMPOSITION OPERATORS(H. Takagi and K. Yokouchi). Still, it just says, "For the proof, refer to Some spectral properties of multipliers on semi-prime Banach algebras(P. Aiena) or Functional analysis(H. Heuser)."

If you know the proof, please help me.