Why do we consider Lebesgue spaces for $p$ greater than and equal to $1$ only? Why do we consider Lebesgue spaces for $p$ greater than and equal to $1$ only and not for $p$ any real number?
 A: Even worse. You can define a $L^p$-distance but the space is not locally convex. See "Examples of spaces lacking local convexity" in http://en.wikipedia.org/wiki/Locally_convex_topological_vector_space.
A: As a concrete example, if we consider the domain $[0,2]$ and try to set $p=1/2$, then our "norm" would be
$$ \|f\|_{1/2} = \left( \int_0^2 \sqrt{|f(x)|} \, dx\right)^2 $$
Consider then
$$ f(x) = \begin{cases} 1 & x<1 \\ 0 & x\ge 1 \end{cases} 
\qquad g(x) = \begin{cases} 0 & x<1 \\ 1 & x\ge 1 \end{cases}$$
We then have $\|f\|_{1/2} = \|g\|_{1/2} = 1$ but $\|f+g\|_{1/2} = 4$, violating the triangle inequality.
A: Mainly because
$$
\Bigl(\int_D|f|^p\Bigr)^{1/p}
$$
is not a norm if $0<p<1$.
A: As it is already mentioned, 
$$
\|f\|_p=\left(\int_X |f|^p\,dx\right)^{1/p},
$$
is a norm iff $p\ge 1$, since only for $p\ge 1$ satisfies the triangle inequality.
Nevertheless, the spaces $L^p(X)$, for $p\in (0,1)$ are quite interesting since they serve as the most typical example for non locally convex topological vector spaces.
A: The reason is that we would like to define a norm by the following formula:
$$
\|f\|_p:=\left(\int_X |f|^p d\mu \right)^{1/p}.
$$
Therefore, we need to have triangle inequality (Minkowski's Inequality) which is available only for $p\geq 1$. See more details in Chapter $6$ of Folland's book: Real Analysis.
