For $p\neq q$ and $p,q\geq 1$, $L^p\neq L^q$

Question

Let $(X,S,\mu)$ be a measure space, and $p\geq 1$, then $L^p:=\{$measurable function $f\mid (\int |f|^p)^{\frac {1}{p}}<\infty\}$

My question is, why isn't $L^p=L^q$ for any $q\geq 1$? Since $f\in L^p \iff (\int |f|^p)^{\frac {1}{p}}<\infty \iff (\int |f|^p)<\infty \iff |f|^p<\infty$ almost everywhere $\iff |f|<\infty$ almost everywhere.

Is there an explanation for why/where my argument was wrong?

Answer 1

It is very false that $\int |f|^p<\infty$ iff $|f|^p<\infty$ almost everywhere. The forward implication is true, but the reverse implication is not. For instance, consider what happens if $f$ is the constant function $1$ on $\mathbb{R}$ with Lebesgue measure. Or, if you want an example where $\mu(X)$ is finite, consider the function $f(x)=1/x$ on $[0,1]$ (with $p=1$).

(Incidentally, it is not always true that $L^p\neq L^p$ if $p\neq q$. For instance, if $X$ is empty, then it is rather trivial that $L^p=L^q$ for all $p$ and $q$.)

Answer 2

There are two monotonicity theorems for $L^p$ spaces:

1. If $X$ has finite measure, then $L^p \subset L^q$ for all $p>q$.

2. If $\inf \{ m(A) : 0<m(A)<\infty \}>0$, then $L^p \subset L^q$ for all $p<q$.

The idea in $1$ is that failure to be in $L^p$ in this case is caused by localized singularities (because long tails cannot exist), and lower powers dampen out localized singularities. The classic example here is $f : (0,1] \to \mathbb{R},f(x)=x^r$ for $r<0$. Here the domain has the Lebesgue measure.

The idea in $2$ is that failure to be in $L^p$ in this case is caused by long tails (because localized singularities cannot exist), and higher powers dampen out long tails. The classic example here is $f : [1,\infty) \to \mathbb{R},f(x)=x^r$ where again $r<0$ and the domain has the Lebesgue measure.

If neither $1$ nor $2$ holds, then there can even be functions in exactly one $p$ with $1<p<\infty$. If both $1$ and $2$ hold then all $L^p$ spaces coincide. A simple example of the latter is the counting measure on a set with $n$ elements; in this case $L^p$ is isometrically isomorphic to $\mathbb{R}^n$ with the $p$ norm (and thus is in particular finite dimensional).

Another interesting result in this general area is that $\{ p<\infty : f \in L^p \}$ is always a closed interval (including the possibility of being empty or a singleton).