$L^p$ space inclusion when $\Omega$ does not contain sets of arbitrarily small measure

Question

I am trying to prove the following:

Let $(\Omega, \Sigma, \mu)$ be a measure space. Assume there exists an $a >0$ such that for any $A \in \Sigma$ either $\mu(A) = 0$ or $\mu(A) \geq a$. Then $L^{p'} \subset L^p$ if $1 \leq p' \leq p \leq \infty$.

I am trying to prove it for simple functions, since the rest would follow easily. To that end, let $f$ be simple. That is $f = \sum_{i = 1}^n \alpha_i 1_{A_i}$, for disjoint $A_i \in \Sigma$. Then,

\begin{align} \|f\|_p &= \|\sum_{i = 1}^n \alpha_i 1_{A_i}\|_p \leq \sum_{i = 1}^n \|\alpha_i 1_{A_i}\|_p \\&= \sum_{i = 1}^n \alpha_i \mu(A_i)^{1/p} \\&= \sum_{i = 1}^n \alpha_i \mu(A_i)^{1/p - 1/p' + 1/p'} \\&\leq \frac{1}{a^{1/p' - 1/p}} \sum_{i = 1}^n \alpha_i \mu(A_i)^{1/p'}. \end{align} The first inequality follows from Minkowski's inequality. Now it is clear that the last term is equal to $\frac{1}{a^{1/p' - 1/p}} \sum_{i = 1}^n \|\alpha_i 1_{A_i}\|_{p'}$. But I am trying to bound $\|f\|_p$ by $\|f\|_{p'}$, so I need to get the sum back into the norm. I am not sure how to continue here, or even if this approach is correct. The exercise this is based on is Exercise 4.1.2 in Achim Klenke's Probability Theory textbook. My approach is based on the proof of the third theorem here (https://math.stackexchange.com/a/66038/697506), but the author does not really explain the last inequality.

Many thanks in advance!

Answer 1

The desired inequality is wrong. Assume $\mu(A_i)=1$ and $\alpha_i=1$ for $i=1\dots n$. Then $$ \|f\|_p = n^{1/p} $$ and $$ \sum_i \alpha_i \mu(A_i)^{1/p'}= n $$ which is larger than $\|f\|_{p'}$ if $p'>1$.

Answer 2

I solved this as follows.

First suppose $p = \infty$, $1\leq p' < \infty$ and $f \in L^{p'}(\mu)$. Define $A_n = \{|f|>n\}$ for $n = 1, 2,...$.

Then, $\int |f|^{p'} 1_{A_n} \mathrm{d}\mu \geq \int n^{p'}1_{A_n} \mathrm{d}\mu = n^{p'}\mu(A_n)$ for all $n$. Thus, $n^{-p'}\int |f|^{p'}\mathrm{d}\mu \geq n^{-p'}\int |f|^{p'} 1_{A_n} \mathrm{d}\mu \geq \mu(A_n) \to 0$ as $n \to \infty$. Since by assumption $\mu(A) = 0$ or $\mu(A) \geq a$ for all $A \in \Sigma$, there exists an $N$ such that $\mu(A_n) = 0$ for all $n \geq N$. So, by definition of $A_n$, $|f| \leq N$ almost everywhere, which implies $\|f\|_{\infty} < \infty$, i.e. $f \in L^\infty(\mu)$.

Now suppose $p, p' \in [1, \infty)$, $p' < p$ and $f \in L^{p'}(\mu)$. Let $A = \{|f|\leq 1\}$ and $B = \{|f| > 1\}$. Then, since $f \in L^{p'}(\mu)$,

$\mu(B) = \int 1_B \mathrm{d}\mu < \int |f|^{p'}1_B \mathrm{d}\mu < \infty$,

and

$\int |f|^p 1_B \mathrm{d}\mu \leq \int \|f\|_{\infty}^p 1_B \mathrm{d}\mu = \|f\|_{\infty}^p \mu(B) < \infty.$

This implies,

$\int |f|^p \mathrm{d}\mu = \int |f|^p 1_A \mathrm{d}\mu + \int |f|^p 1_B \mathrm{d}\mu \leq \int |f|^{p'} 1_A \mathrm{d}\mu + \int |f|^p 1_B \mathrm{d}\mu < \infty$.

Thus, $f \in L^p(\mu)$, and also in this case we have the inclusion.

Answer 3

We know that $\mu(\{x\in X:|f|>n\})=0$ for some sufficiently large $n$ by the supposition, Chebyshev's inequality, and the fact that $f\in L^{p'}$. So $f/n$ is less than $1$ outside a null set, from which we conclude that $\| f/n \|_p\leq \| f/n \|_{p'}$, which gives the result.

Edit: Here are the details: suppose $f\in L^{p'}$. Then $\|f\|_{p'}<\infty \implies \mu(\{x\in X:|f|>n\}) \leq n^{-p'} \|f\|_{p'}$ goes to $0$ as $n\to \infty$, because $p'\geq 0$. In particular, for large enough $n$, $\mu(\{x\in X:|f|>n\}) <a\implies \mu(\{x\in X:|f|>n\}) = 0$ since there are no sets of positive measure less than $a$ in the space. So $\mu(\{x\in X:|f/n|>1\})=0$; let $A=\{x\in X:|f/n|>1\}$ so that $X=A\cup A^c$ with $\mu(A)=0$. Since $1\leq p'\leq p$, $x\in A^c \iff|f(x)/n|\leq 1 \implies |f(x)/n|^p \leq |f(x)/n|^{p'}$. Therefore, $\|f\|_p\leq \|f\|_{p'}<\infty \ \implies f\in L^p$.