$L^q$ contains $L^p$ but $l^p$ contains $l^q$? Suppose $1\leq q < p < \infty$. 
For finite Lebesgue measure sets $X$ ($\mu(X) < \infty$), we have
$$
L^p(X) \subset L^q(X)
$$
On the other side, for the infinite counting measure $\mu$ ($\mu(\mathbb{N})=\infty$), we have
$$
L^q(\mathbb{N}) = l^q \subset l^p = L^p(\mathbb{N})
$$
Why does the containment reverse directions? Is is because they use different measures, because one measure is finite and the other infinite, or something else? I'd like to have an intuitive understanding why this reversal occurs.
 A: For $L^p(X)$, where $X$ is a space with finite total measure that can be subdivided finely, the integral is bad when the function becomes very large on some narrow slice. For example, consider improper integrals $\int_0^1 f(x)\,dx$ with $\lim_{x\to 0} f(x)=\infty$; it's all about exactly how fast that happens.
For these, with large values on subsets of small measure, taking higher powers makes things blow up faster, and $L^p\subset L^q$ when $p>q$.
For sequence spaces $\ell^p$, our sum is an integral with respect to counting measure - infinite total measure, but we can't slice it up into small pieces. Here, the sum can only work if most values are $<1$, and it goes bad if they don't decay fast enough.
Higher powers make these decay faster, and $\ell^q\subset \ell^p$ when $q<p$.
If the measure space has both qualities - infinite total measure, non-atomic - such as $\mathbb{R}$, the integral can go bad in either way, and neither space is contained in the other.
If the measure has neither quality - finite total measure, a lower limit on the size of non-null sets - then it might as well be counting measure on a finite set. The two spaces are equal, because they're both as large as they could possibly be.
