Consider a general measure space $(\mu, \Sigma,X)$ with $\mu(X)=1$. I want to show that for any $\mu-$measurable function and for any $0<p<\infty$, we have $$\|f\|_{p}\leq \|f\|_{\infty}.$$
A similar question is here Show $||f||_p\leq ||f||_\infty$, but it is in an easier case, in the space $C([0,1])$, it seems that the $\sup-$norm is the same as the $\sup_{x}|f(x)|$. However, in a general measure space, the $\sup-$norm is actually an essential $\sup$: $$\|f\|_{\infty}=\inf\{M\geq 0:|f(x)|\leq M\ \text{holds for almost every}\ x\}.$$
Thus, the only thing I know is that $|f(x)|\leq\|f\|_{\infty}$ $\mu-$almost surely. I want to start from here but I got stuck:
We have $|f(x)|^{p}\leq\|f\|_{\infty}^{p}$ $\mu-$almost surely, and thus the set defined by $E:=\{x\in X:|f(x)|^{p}>\|f\|_{\infty}^{p}\}$ has $\mu(E)=0$.
Now, $$\int_{X}|f^{p}(x)|\mu(dx)\leq \int_{X\setminus E}|f(x)|^{p}\mu(dx)+\int_{E}|f(x)|^{p}\mu(dx).$$
The first integral is easy to evaluate, since $\mu(X)=1$, we know that $\mu(X\setminus E)=1$, and thus $$\int_{X\setminus E}|f(x)|^{p}\mu(dx)\leq \int_{X\setminus E}\|f\|_{\infty}^{p}\mu(dx)=\|f\|^{p}_{\infty}\mu(X\setminus E)=\|f\|^{p}_{\infty}.$$
But how to evaluate the second integral? Over $E$ I just have a reversed directly $>$. I have no idea how to bound it..
Thank you for any help!!
Okay. Sorry for the stupid question. As the comment suggested, the integral of any measurable function over a zero measure set is $0$.
Indeed, one could see here Integral over null set is zero.