When $L^p \subset L^q$ for $p

Question

Assume that $(X, \mathfrak{B}, m)$ is a measure space such that there exists a constant $\alpha>0$ such that for every $E \in \mathfrak{B}$ the following holds: $$ m(E)=0 \ \ or \ \ m(E)\geq \alpha.$$

Is it then true that for every $1\leq p \leq q \leq \infty$ $$L^p((X, \mathfrak{B}, m) \subset L^q(X, \mathfrak{B}, m)?$$

I know that it is true for spaces $l^p$ (it is a particular case of $L^p$ when $X=\mathbb{N}$, $\mathfrak{B}=2^\mathbb{N}$ and $m$ is a counting measure) -proof is for example here How do you show that $l_p \subset l_q$ for $p \leq q$?.

My question is related to the last theorem in the another answer https://math.stackexchange.com/a/66038/20924 . Here is its proof, but not all is clear for me. I have one doubt. It seems that here the following equalities are used: $$\|f\|_{L^p}=\sum_{j=1}^n a_j m(E_j)^{1/p},$$ $$\|f\|_{L^q}=\sum_{j=1}^n a_j m(E_j)^{1/q}$$ for $f(x)=\sum_{j=1}^n a_j \chi_{E_j}$, where $E_j$ are pairwise disjoint, which are generally not not true even for $l^p$ and $l^q$.

Answer 1

Let $f\in L^p(X,\mathfrak B,\mu)$. We define $A_n:=\{x\in X: n\leq |f(x)|<n+1\}\in\mathfrak B$. We have $\sum_{n\geq 0}\int_{A_n}|f|^pd\mu<\infty$ so $\sum_{n\geq 0}n^p\mu(A_n)<\infty$. By the hypothesis on $\mu$, we have for $n$ large enough, say $n\geq n_0$, that $\mu(A_n)=0$ hence $\lvert f\rvert\leqslant n_0+1$. So if $q=\infty$ we are done and if $q<\infty $ we have $$ \lvert f(x)\rvert^q=\lvert f(x)\rvert^p \lvert f(x)\rvert^{q-p}\leqslant\lvert f(x)\rvert^p (n_0+1)^{q-p}.$$

Answer 2

Thanks to Davide Giraudo's idea, one can consider an alternative solution using Chebyshev's inequality.

Define the sets $$ A_n := \{x \in X : n \leq |f(x)|^p < n+1\}, \quad \text{for } n = 0, 1, 2, \ldots $$

By Chebyshev's inequality, we have $$ \mu(A_n) \leq \frac{\|f\|_{L^p}^p}{n} \to 0 \quad \text{as } n \to \infty. $$

However, by the assumption on the measure $\mu$, we know that it is either identically zero or uniformly bounded above away from zero on sets of positive measure. Since $$ \mu(A_n) \to 0, $$ This implies that $$ \mu(A_n) = 0 \quad \text{for sufficiently large } n. $$

Therefore, $$ |f(x)|^p < N \quad \text{almost everywhere for some large enough } N, $$ which means that $f$ is essentially bounded.

The case $q > p$ is the same as in Davide Giraudo's solution.