Let $(X,\mathcal{A}, \mu)$ be a measure space and there exists a $p_{0}$ so that $f \in \bigcap\limits_{p \in [p_{0}, \infty[}L^{p}$ and further $\sup\limits_{p \in [p_{0}, \infty[}\vert \vert f \vert \vert_{p}<\infty$.
Show that $f \in L^{\infty}$.
Questions: Can someone explain to me why the statements $f \in \bigcap\limits_{p \in [p_{0}, \infty[}L^{p}$ and $\sup\limits_{p \in [p_{0}, \infty[}\vert \vert f \vert \vert_{p}<\infty$ are not equivalent? It most probably has something to do with the definition of $\sup$ but I do not see the difference. Further, any ideas on how I can show $f \in L^{\infty}$. I am tending towards a proof via cotradiction. So, let's assume $f \in L^{\infty}$, it then follows that for any $\alpha> 0$ there exists an $A\in \mathcal{A}$ so that $|f(a)|>\alpha$ for all $a \in A$. And further $\mu(A) > 0$. How would this help me?
If $\|f\|_\infty=\infty$, then for every $M>0$, we would have that $$ \mu(E_M)>0,\quad\text{where $E_M=\{x\in X: |f(x)|\ge M\}$}. $$ [Note that $\|f\|_\infty$ is the essential supremum of $|f(x)|$.]
Hence, for every $p\ge p_0$, $$ \|f\|_p=\left(\int_X|f|^p\,d\mu\right)^{1/p} \ge\left(\int_{E_M}|f|^p\,d\mu\right)^{1/p} \ge\left(\int_{E_M}M^p\,d\mu\right)^{1/p}=\mu^{1/p}(E_M)\cdot M $$ This implies that $$ \lim_{p\to\infty}\|f\|_p\ge \lim_{p\ge p_0} \mu^{1/p}(E_M)\cdot M=M. $$ Hence, if $\|f\|_\infty=\infty$, then we would also have that $\lim_{p\to\infty}\|f\|_p=\infty$.