$L^p$ and $L^q$ spaces do not include each other.

Asked Feb 03 '13 at 01:49

Active Feb 03 '13 at 02:21

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According to the answer at this question, the following statement should be true.

Let $(X, \mathcal{A}, \mu)$ be a measure space without atoms and such that $\mu(X)=\infty$. Then, $L^p(\mu)$ and $L^q(\mu)$ do not contain each other.

Where can I find a proof of this assertion?

edited Apr 13 '17 at 12:20

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asked Feb 03 '13 at 01:49

Amanda Seyfried

2

In order for this to be true, you'll also want to say that $\mu(X)$ is infinite... – Micah Feb 03 '13 at 01:56
7

Loved you in Mamma Mia – Will Jagy Feb 03 '13 at 01:59
I guess that a way of proving this (with the additional hypothesis $\mu(X)=\infty$, thanks @Micah) is to show that a measurable function $F\colon X\to \mathbb{R}$ exists such that $$\mu(F^{-1}(A))=\mathcal{L}^1(A),$$for all Borel subset $A$ of $\mathbb{R}$. For something related to this, see this question and espectially this answer by Byron Schmuland[...] – Giuseppe Negro Feb 03 '13 at 02:01
@micah : Thanks. – Amanda Seyfried Feb 03 '13 at 02:03
Once this is done, it will suffice to take a function $f\in L^p(\mathbb{R})$ such that $f\notin L^q(\mathbb{R})$ and pull it back via $F$. By the change of variable formula we will have $f\circ F \in L^p(X),\ f\circ F \notin L^q(X)$. In an analogous way you disprove the opposite inclusion. – Giuseppe Negro Feb 03 '13 at 02:04

$L^p$ and $L^q$ spaces do not include each other.

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