Proof verification: $fg\in L^1$ for any $g\in L^q$ implies that $f\in L^p$ This is a classical exercises. I know its proof relying on constructions. 
What below is my attempt from the view point of functional analysis. 

Let $(X,\mu)$ be a $\sigma$-finite measure space. Let $1\leq q<\infty$ and $p=\frac{q}{q-1}$. 
  If $fg\in L^1$ for any $g\in L^q$, then $f\in L^p$. 

Consider the linear operator $T_f(g):=fg$ from $L^q$ into $L^1$. We want to show that $T_f$ is bounded. i.e.,
$$
\|T_f(g)\|_{L^1}\leq C\|g\|_{L^p}
$$
for some $C>0$.
To see this, we show that it is a  closed operator defined on $L^q$. To this end, we only need to show for any convergent sequences  $u_n\to u$ in $L^p$  and $T_f(u_n)\to y$ in  $L^1$, it holds that $T_f(u)=y$. 
To see this, we can select a subsequences  $u_{n_k}$ such that $u_{n_k}\to u$ a.e. and $T_f(u_{n_k})=f u_{n_k}\to y$ a.e. Then we have $f=gh$ a.e.
Therefore, $T$ is a closed operator and hence it is bounded by closed graph theorem. 
This shows that $g \mapsto \int_X fg d\mu$ is also a bounded linear functional on $L^q$. By duality, the norm of this functional  is the $L^p$ norm of $f$. Hence $\| f \|_p \le C < \infty.$ Thus $f \in L^p$. 
[Updated proof:]

Consider the linear operator $T_f(g):=fg$ from $L^q$ into $L^1$. We want to show that $T_f$ is bounded. i.e.,
  $$
\|T_f(g)\|_{L^1}\leq C\|g\|_{L^q}
$$
  for some $C>0$.
  To see this, we show that it is a  closed operator defined on $L^q$. To this end, we only need to show for any convergent sequences  $u_n\to u$ in $L^p$  and $T_f(u_n)\to y$ in  $L^1$, it holds that $T_f(u)=y$. 
To see this, we can select a subsequences  $u_{n_k}$ such that $u_{n_k}\to u$ a.e. and $T_f(u_{n_k})=f u_{n_k}\to y$ a.e. The process goes as follows: we first choose a subsequence $u_{n_k}$ such that  $u_{n_k}\to u$ a.e.. Then we choose a subsequence of $u_{n_k}$, which still denoted by $u_{n_k}$, such that 
  $fu_{n_k}\to y$ a.e. Then it is obvious that $y=fu$ a.e. Indeed, there exit sets $N_1,N_2$ of measure zero such that $fu_{n_k}(x)\to y(x)$ over $X\backslash N_1$ and $u_k(x)\to u(x)$ over $X\backslash N_2$. Then $y(x)=f(x)u(x)$ over the set $X\backslash (N_1\cup N_2)$
Therefore, $T$ is a closed operator and hence it is bounded by closed graph theorem. 
This shows that $g \mapsto \int_X fg d\mu$ is also a bounded linear functional on $L^q$. By duality, the norm of this functional  is the $L^p$ norm of $f$. Hence $\| f \|_p \le C < \infty.$ Thus $f \in L^p$. 

 A: proof-verification

Consider the linear operator $T_f(g):=fg$ from $L^q$ into $L^1$. We want to show that $T_f$ is bounded. i.e.,
  $$
\|T_f(g)\|_{L^1}\leq C\|g\|_{L^p}
$$
  for some $C>0$. (Typo: p should be q. I'm sure you know C does not depend on g, but I would be a bit careful with the quantifier here)
  To see this, we show that it is a  closed operator defined on $L^q$. To this end, we only need to show for any convergent sequences  $u_n\to u$ in $L^p$  and $T_f(u_n)\to y$ in  $L^1$, it holds that $T_f(u)=y$. 
  To see this, we can select a subsequences  $u_{n_k}$ such that $u_{n_k}\to u$ a.e. and $T_f(u_{n_k})=f u_{n_k}\to y$ a.e. (Why? Sure, L^p convergence implies existence of almost everywhere convergent subsequence. But you are dealing with the sequence T_f(u_n) as well. Elaboration?)Then we have $f=gh$ a.e. (Your proof breaks down here. Typo? You are supposed to show fu = y. Why is this true?)
  Therefore, $T$ is a closed operator and hence it is bounded by closed graph theorem. 
This shows that $g \mapsto \int_X fg d\mu$ is also a bounded linear functional on $L^q$. By duality, the norm of this functional  is the $L^p$ norm of $f$. Hence $\| f \|_p \le C < \infty.$ Thus $f \in L^p$. 

