$fg\in L^1$ for all $g\in L^q$ $\Longrightarrow f\in L^p$

Question

Let $(X,\mathscr{M},\mu)$ be a $\sigma$-finite measure space and $f$ be a $\mathscr{M}$-measurable function. Let $p,q$ be Hölder conjugates of each other where $1\leq p\leq\infty$. Then is it true that

$fg\in L^1$ for all $g\in L^q$ $\Longrightarrow f\in L^p$ ?

I've managed to prove this for $1\leq p<\infty$. I argued as follows:

First, it is easy to check that $f$ must be finite almost everywhere. Now define $\nu(A)=\int_A |f|^p \,d\mu$ for each $A\in\mathscr{M}$. Then it is also easy to check that $(X,\mathscr{M},\nu)$ is a $\sigma$-finite measure space, and the assertion is that this is actually a finite measure space. Assume $\nu(X)=\infty$ for contradiction. Then there exist a measurable function $h$ such that $$h\in L^p(X,\mathscr{M},\nu)\text{ for all }p>1,\text{ but }h\notin L^1(X,\mathscr{M},\nu).$$ Then the function $g=h|f|^{p-1}$ is in $L^q(X,\mathscr{M},\mu)$ but we have $fg\notin L^1(X,\mathscr{M},\nu)$, a contradiction.

The case $p=\infty$ must be handled separately, but I don't see how I should proceed...

This is an exercise from Jones' Lebesgue Integration on Euclidean Space, and the book does not contain the theory of Banach spaces, dual spaces, Riesz representation theorem, etc. The proof I'm looking for is therefore the one avoiding such advanced theories. Can someone show me how to handle the case $p=\infty$ in a (relatively) elementary manner? Any advice is welcome. Please enlighten me.

Answer 1

Suppose $f$ is unbounded. We would like to show that there exists nonnegative $g\in L^1$ such that $f\cdot g \notin L^1$. We may assume $f$ is nonnegative and we may replace $f$ with a smaller function such that the following is true: There exists a sequence of disjoint sets of positive measure $\{A_k\}_{k\in\mathbb{N}}$ such that $k\leq f(x)<k+1$ whenever $x\in A_k$. Now for $x\in A_k$ define $g (x):=\frac{1}{k^2\mu(A_k)}$ and define $g(x) := 0$ for $x\notin \cup_{k=1}^\infty A_k$. Then $g$ has the desired properties.