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

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 http://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 $l^q$.

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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 $\mu(A_n)=0$. So if $q=\infty$ we are done and if $q<\infty$ we have $$\int_{A_n}|f|^qd\mu\leq \mu(A_n)n^q,$$ which is the general term of a convergent series since it's equal to $0$ for $n\geq n_0$. So $\int_X|f|^q<\infty$ and $f\in L^q(X,\mathfrak B,\mu)$.