(X, sigma, measure) is a finite measure space. Show that as p--> infty, lim |f|_p = |f|_infty, where f lies in L^infty.

I know that

|f|_infty= ess sup |f|= smallest number k st measure ({x:|f||>k})=0

I would note that given e>0, measure({x in X: |f(x)|> |f|_infty - e} ) > 0.

But from here, I am kind of lost.