$\mathcal{L^p}$ norm induces a norm on quotient space Let $(S, \Sigma, \mu)$ be a measure space. We know that $||. ||_p$ is not a proper norm on $\mathcal{L}^p(S, \Sigma, \mu)$ because $||f||_p=0 $ only implies $f=0$ a.e.
The measure theoretic probability text I am reading then goes on by defining the equivalence relation $f\sim g$ iff $\mu(\{f\neq g\})=0.$ It is claimed that the norm $||.||_p$ induces a norm on the quotient space $L^p:=\mathcal{L}^p/ \sim $. From what I understand, this quotient space is simply a partition of $\mathcal{L}^p$.
I lack any knowledge about functional analysis$-$ I am wondering how the 'induced' norm on $L^p$ is defined.
 A: Write $[f]$ the equivalent class of $f$ and define $\|[f]\|_{L^{p}}=\left(\displaystyle\int|f|^{p}d\mu\right)^{1/p}$, one can show that any representative of $f$, say, $f=g$ a.e. then $\left(\displaystyle\int|f|^{p}d\mu\right)^{1/p}=\left(\displaystyle\int|g|^{p}d\mu\right)^{1/p}$, so the definition of $\|[f]\|_{L^{p}}$ is okay. Now if $\|[f]\|_{L^{p}}=0$, then $\displaystyle\int|f|^{p}d\mu=0$, as $|f|^{p}\geq 0$, it is a standard fact that $|f|=0$ a.e. then $f=0$ a.e. so $[f]=[0]$.
A: The point is that $||f||_p$ is not a norm on functions because, as you say, $||f||_p = 0$ would imply $f$ identically $0$. However, if you work for example on $\mathbb{R}$, the function
$$f \colon \mathbb{R} \rightarrow \mathbb{R}, f(0) = 1, f(x) = 0 \text{ when } x \neq 0$$
satisfies $||f||_p = 0$, and it is not identically $0$! So one identifies functions which coincide almost everywhere, so that $f$ in my example is actually equivalent to the function identically $0$, which has norm $0$. This way you define $|| \cdot ||_p$ on equivalence classes of functions which is a well-defined norm.
