# If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?

Let $f$ be a measurable function on a measure space $X$ and suppose that $fg \in L^1$ for all $g\in L^q$. Must $f$ be in $L^p$, for $p$ the conjugate of $q$? If we assume that $\|fg\|_1 \leq C\|g\|_q$ for some constant $C$, this follows from the Riesz Representation theorem. But what if we aren’t given that such a $C$ exists?

So that $$L^p$$ and $$L^q$$ are duals of each other, let $$1\lt p,q\lt\infty$$, as well as $$\frac1p+\frac1q=1$$.
Suppose that $$fg\in L^1$$, but there is no $$C$$ so that $$\|fg\|_{L^1}\le C\|g\|_{L^q}$$. Without loss of generality, we can assume all functions are positive. Suppose we have a sequence of $$L^q$$ functions $$\{g_k:\|g_k\|_{L^q}=1\}$$ where $$\int|fg_k|\;\mathrm{d}x>3^k$$. Set $$g=\sum\limits_{k=1}^\infty2^{-k}g_k$$. $$\|g\|_{L^q}\le1$$ yet $$fg\not\in L^1$$. Thus, there must be a $$C$$ so that $$\|fg\|_{L^1}\le C\|g\|_{L^q}$$. Then, as you say, apply the Riesz Representation Theorem.