Construct a measure space s.t. $f\in L^p(X)\rightarrow f\in L^1(X)$ I want to construct a measure space s.t.$\mu(X)=\infty$ and $f\in L^p(X)$ for some $1<p<\infty\rightarrow f\in L^1(X)$
 A: Take any uncountable set $X$ and let the measure $\mu$ be defined on the countable/co-cocountable sigma-algebra of $X$ by $\mu(A)=0$ if $A$ is countable and $\mu(A)=\infty$ if $X\setminus A$ is countable. Then $L^p(X,\mu)=\{ 0\}$ for any $p\in [1,\infty)$, so your implication holds. To see that $L^p(X,\mu)=\{ 0\}$ for all $p$, it is enough to 
do this for $p=1$. Consider any $f\in L^1(X,\mu)$. Then $\mu(\{ \vert f\vert\geq \varepsilon\})<\infty$ for every $\varepsilon >0$ by Markov's inequality, so $\mu(\{ \vert f\vert\geq\varepsilon\})=0$ by the definition of $\mu$. Since this holds for every $\varepsilon >0$, it follows that $f=0$ almost everywhere.
On the other hand, the implication does not hold as soon as $X$ contains (measurable) sets with arbitrarily large finite measure. Indeed, in this case one can find a sequence of pairwise disjoint measurable sets $A_n$ with $1\leq \mu(A_n)<\infty$ for all $n$. (Start with $A_1$ such that $1\leq \mu(A_1)<\infty$; then choose $B_2$ such that $2\leq \mu(B_2)<\infty$ and set $A_2:=B_2\setminus A_1$; etc). Set $m_n:= \mu(A_n)$. Given $p>1$, choose any sequence of positive numbers $(\alpha_n)$ such that $\sum_1^\infty m_n \alpha_n^p<\infty$ and $\sum_1^\infty m_n\alpha_n=\infty$. (This can be done because $\inf_n m_n>0$: take any sequence $(\beta_n)$ such that $\sum \beta_n^p<\infty$ but $\sum\beta_n=\infty$, and set $\alpha_n:=m_n^{-1/p}\beta_n$). Define $f:X\to [0,\infty)$ by $f\equiv \alpha_n$ on $A_n$ and $f\equiv 0$ outside $\bigcup_n A_n$. Then $f$ is in $L^p(X,\mu)$ but not in 
$L^1(X,\mu)$.
Conversely, if $L^p(X,\mu)$ is not contained in $L^1(X,\mu)$ for some $p>1$, then $X$ contains sets of arbitrarily large finite measure. Indeed, take any $f\in L^p\setminus L^1$. For any $\varepsilon>0$, the set $A_\varepsilon:=\{ \vert f\vert\geq\varepsilon\}=\{ \vert f\vert^p\geq\varepsilon^p\}$ has finite measure by Markov's inequality. Moreover, the set $A=\{ \vert f\vert>0\}=\{ f\neq 0\}$ has infinite measure because $L^p(A)$ is not contained in $L^1(A)$. So we have $\lim_{\varepsilon\to 0}\mu(A_\varepsilon)=\infty$.
