When $L_p = L_q$? As we know that $L_p \subseteq L_q$ when $0 < p < q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks!
 A: Equality only when $p = q$. Suppose $p < q$, and let $r = \frac{p+q}{2}$. 
Let $B_n$ be a sequence of disjoint measurable subsets of your probability space with the property that their measures $|B_n| = m_n \leq 2^{-n}$. 
Let $\chi_n$ denote the characteristic function of $B_n$. Consider the sequence of functions 
$$f_n = \sum_1^n \frac{1}{m_k^{1/r}} \chi_k$$
It is easy to check that this sequence is bounded in $L^p$ and Cauchy, so converges to a limit function $f$. But the sequence is unbounded in $L^q$. Hence the limit function $f$ is an element of $L^p$ that is not in $L^q$. 
A: For probability measure $ L_p \subset L_q $ .
Equality is unreachable, see http://en.wikipedia.org/wiki/Lp_space. 
A: This is essentially Exercise 6.5 in Folland's Real Analysis.  Let $(X,\mathcal{F},\mu)$ be a measure space, and let $$m = \inf\{\mu(A) : A \in \mathcal{F}, \mu(A) > 0\}$$
$$M = \sup\{\mu(A) : A \in \mathcal{F}, \mu(A) < \infty\}$$
For $0 < p < q < \infty$, it is a fact that $L^p(\mu) \subset L^q(\mu)$ iff $m > 0$, and $L^q(\mu) \subset L^p(\mu)$ iff $M < \infty$ (which in particular holds when $\mu$ is finite).
So for a finite measure $\mu$, it is necessary and sufficient that for some $m >  0$ every set either has measure 0 or has measure at least $m$.  This is going to force your space to be "effectively" finite in some sense.
