Measure spaces such that the semi-norm is a norm In my lecture notes there is the following exercise:
"Characterize those measure spaces $(X, B, \mu)$ on which the semi-norm $\|f\| = \int_X |f| d \mu$
defined on $L^1(X) = \{ f \mid f \text{ measurable and} \int_X |f| d \mu < \infty \}$ of Lebesgue-integrable functions is a norm."
I thought that if I take $X$ to be finite and $\mu$ to be the counting measure then $\|\cdot\|_1$ is a norm. But I think the exercise asks me to use the Lebesgue measure so $\mu(X) = 0$ if $X$ is finite and my example breaks.
What's the correct answer?
 A: Assume that $\lVert \cdot \rVert$ is a norm. If $N$ is a measurable set of measure $0$, then $\int_X\chi_Nd\mu=0$ hence $\chi_N$ is identically $0$ and $N$ is the emptyset. So each non-empty measurable set set has a positive measure. 
Conversely, if each non-empty measurable set has a positive measure, let $f$ a function such that $\lVert f\rVert=0$. Then 
$$0=\lVert f\rVert\geq\int_{\{f\geq 2^{-n}\}}|f|d\mu\geq 2^{—n}\mu(\{|f|\geq 2^{-n}\})$$ 
hence $\mu(\{|f|\geq 2^{-n}\})=0$ and $\mu(\{|f|>0\})=0$. By hypothesis, it implies that $\{|f|>0\}$ is empty, hence $f=0$. 

In general, to define a norm by this way, we rather consider the equivalence classes of functions. Otherwise, we may have some problems. For example, if $\mathcal B$ is the smallest $\sigma$-algebra containing all the open sets of a Hausdorff topology, each singleton is measurable and should have a positive measure. So $X$ can only be at most countable. 
A: Here is a large source of them.  Suppose $\Omega$ is a set that is partitioned by subsets
$$\{E_n\}_{n=1}^\infty$$
Suppose that $\mu$ is a measure on the $\sigma$ algebra $\mathcal{S}$ generated by the $E_k$ and that $\infty > \mu(E_k) > 0$, $1\le k \le n$.  Then $f\mapsto \int_{\Omega} |f|\,d\mu$ is a norm on $L^1(\Omega, \mathcal{S}, \mu)$.  
Some variation on this is probably your complete answer.  Note that any $\mu$-null set that is nonvoid does you in on the spot.
