Is the $p$-norm ever a norm for $0I wonder: 
Is there a measure space $(X,\Sigma,\mu)$ such that $L^p(\mu)$ form a normed space w.r.t the $p$-norm, for some $0<p<1$?(assuming that $X$ contains more than point).
I know that in general the "$p$-norm" is not really a norm for $0<p<1$. 
(it violates the triangle inequality). I am asking if there are (non-trivial) special cases where it does form a norm.
 A: Just to close this question, I am expanding Yanko's comment into an answer:
Let $(X,\Sigma,\mu)$ be a measure space. Suppose that there exist
measurable sets $A,B$ such that $\mu(A),\mu(B)>0$ but $\mu(A\cap B)=0$.
Set $f=1_A,g=1_B$.
Then, $\|f+g\|_p^p=\int_X (1_A+1_B)^p=\int_{A \cup B} (1_A+1_B)^p=\int_{A\setminus B} (1_A+1_B)^p+\int_{B\setminus A}(1_A+1_B)^p=$
$\int_{A\setminus B} 1_A+\int_{B\setminus A}1_B=\mu(A)+\mu(B)$,and $\|f\|_p=(\mu(A))^{\frac{1}{p}},\|g\|_p=(\mu(B))^{\frac{1}{p}}$.
Thus $$ \|f+g\|_p \le \|g\|_p+\|g\|_p \iff (\mu(A)+\mu(B))^{\frac{1}{p}} \le (\mu(A))^{\frac{1}{p}}+(\mu(B))^{\frac{1}{p}}$$.
When $0<p<1$, we have  $r=\frac{1}{p}>1$, and $(x+y)^r>x^r+y^r$, so the triangle inequality does not hold. Indeed, we can write
$$ (x+y)^r=(x+y)(x+y)^{r-1}=x(x+y)^{r-1}+y(x+y)^{r-1}>xx^{r-1}+yy^{r-1}=x^r+y^r,$$ where in the strict inequality step we have used the fact 
$ (x+y)^{r-1}>\max\{x^{r-1},y^{r-1}\}$, i.e. the fact that the function $x \to x^s$ is strictly increasing for $s>0$.
