About multiplication operator on $L^p(X, \mu)$ where X is sigma finite Let X be sigma finite measure space and $\phi$ $\in$ $L^{\infty}(X, \Omega)$ and $M_\phi:L^p(X, \Omega)$ $\to$ $L^p(X, \Omega)$ multiplication operator then show that $\| M_\phi \|=\| \phi\|_{\infty}$.
My attempt:
I could prove that $\| M_\phi \| \leq \| \phi \|_ {\infty}$ 
And then to prove the reverse inequality, I tried constructing a sequence of functions $f_n$ in $L^p$ such that $\| \phi f_n \|_p$ converges to $\| \phi \|_ \infty$ in the field. If we get such a sequence then that will prove the result by property of sup. Can we construct such a sequence?
As X is sigma finite so $X=\cup_{i=1}^{\infty} X_i$ where $\mu(X_i) < \infty$
I tried defining $f_n(x)=1 $ if $x \in \cup_{i=1}^n X_i$ and $0$ otherwise.
But then the norm of $M_\phi(f_n)$ converges to $\int \phi^p$
Can i modify this? Or is there any other way?

I think i have solved it. Can someone please check it.. 
$\| \phi \|_{\infty} = inf$ $\lbrace c>0 : \mu(\lbrace x\in X:|\phi(x)|>c\rbrace)=0\rbrace$
let $C_X=\big\lbrace c>0 : \mu(\lbrace x\in X:|\phi(x)|>c\rbrace)=0\big\rbrace$
now clearly $\| \phi \|_{\infty}- \epsilon \notin C_X$ So $\mu(\lbrace x\in X:|\phi(x)|>\| \phi \|_{\infty}- \epsilon \rbrace)\neq 0$
And using sigma finiteness of X from the above set we can get a set D whose measure is finite non zero and on which $|\phi(x)|>\| \phi \|_{\infty}- \epsilon $.   Then define $f= \dfrac{\chi_D}{\mu(D)^{1/p}}$ Then for this $f$ its easy to prove that it is infact in $L^P$ and $\|M_\phi(f)\| \geq \| \phi\|_\infty-\varepsilon$ and letting $\varepsilon$ tends to zero we get the required result.
