if $\mu(X)=1$ , $f\in L^1$, then $f\in L^\infty$?

Question

if $\mu(X)=1$ , $f\in L^1$, then is true that $f\in L^\infty$?

Basically, I was wondering if I can write $$\int_X f d\mu \le \mu(X). \|f\|_\infty < \infty$$

or can we claim $\|f\|_\infty =1$? I'm thinking in the eyes of probability that f can at most take value of 1.

but if yes how to prove it?

Answer 1

No. For example, $f(x) = \frac{1}{\sqrt{x}}$ on $X = (0,1]$ has $f \in L^1 \setminus L^\infty$