Given $f\notin L^p$ find $g\in L^q$ s.t. $fg\notin L^1$

We first show that for every $(a_n)_{n}$ not in $\ell^p$, there exists $(b_n)_{n}$ in $\ell^q$ such that $(a_nb_n)_{n}$ is not in $\ell^1$. To this end, assume without loss of generality that every $a_n$ is positive and introduce the sequences $$ A_n=1+\sum\limits_{k\leqslant n}a_k^p,\qquad b_n=\frac{a_n^{p-1}}{A_n}. $$ To show that $(b_n)_{n}$ is in $\ell^q$, note that $$ (q-1)b_n^q=(q-1)\frac{a_n^{p}}{A_n^q}=(q-1)\frac{A_n-A_{n-1}}{A_n^q}\leqslant\frac1{A_{n-1}^{q-1}}-\frac1{A_{n}^{q-1}}. $$ The last inequality stems from the fact that $1+(q-1)u^q\geqslant qu^{q-1}$ for every $u$ in $[0,1]$, used for the value $u=A_{n-1}/A_n$.

Summing this, one gets $$ (q-1)\sum\limits_{n}b_n^q\leqslant1. $$ To show that $(a_nb_n)_{n}$ is not in $\ell^1$, note that $a_kb_k=a_k^p/A_k=(A_k-A_{k-1})/A_k$ for every $k$ hence, for every $m\geqslant n$, $$ \sum\limits_{k=n+1}^ma_kb_k\geqslant\sum\limits_{k=n+1}^m(A_k-A_{k-1})/A_m=(A_m-A_{n})/A_m. $$ By hypothesis $(A_k)_k$ is unbounded hence, for every $n$, there exists $m\geqslant n+1$ such that $A_m\geqslant2A_n$. For such $n$ and $m$, $\sum\limits_{k=n+1}^ma_kb_k\geqslant\frac12$. This proves that $(a_nb_n)_{n}$ is not in $\ell^1$.

Turning to the case of functions, assume without loss of generality that $f$ is nonnegative and consider the case when $E=\mathbb R_+$ with the Lebesgue measure. Define $F:x\mapsto1+\int\limits_0^xf^p$ and $g=f^{p-1}/F$.

Then $g^q=f^p/F^q$ and $f^p=F'$ hence $$ (q-1)\int_{\mathbb R^+}g^q=(q-1)\int_{\mathbb R^+}\frac{F'}{F^q}=\left[\frac{-1}{F^{q-1}}\right]_0^{+\infty}\leqslant1. $$ Likewise $fg=f^p/F=F'/F$ hence $$ \int_{\mathbb R^+}fg=\left[\log F\right]_0^{+\infty}, $$ which is infinite since $F(x)\to\infty$ when $x\to+\infty$. Thus $g$ is in $L^q$ and $fg$ is not in $L^1$.

The case of functions defined on a general space $E$ can be deduced from this one. Namely, one considers the function $g:x\mapsto f(x)^{p-1}/h(f(x))$ where, for every nonnegative $t$, $$ h(t)=1+\int\limits_Ef^p(y)\,[f(y)\geqslant t]\,\mathrm d\mu(y). $$ Then, if $\mu(f\geqslant t)$ is finite for every positive $t$, $fg$ is not in $L^1$ and $g$ is in $L^q$.