Functions whose product with every $L^1$ function is $L^1$

Question

Let $\mu$ be a probability measure and $f$ a measurable function whose product with any integrable function is integrable: $$ \int|g|\,{\rm{d}}\mu<\infty\implies \int|fg|\,{\rm{d}}\mu<\infty. $$ Does this imply $f\in L^\infty(\mu)$?

Observation 1) $f$ amounts to a well-defined functional $g\mapsto\int fg\,{\rm{d}}\mu$ on $L^1(\mu)$. If one argues that this is bounded, the Riesz representation theorem can be invoked.

Observation 2) Limiting to $g\in\{1,f,f^2,\dots\}$, a simple induction shows that $f\in L^n(\mu)$ for every positive integer $n$. Thus $f$ lies in $\bigcap_{n=1}^{\infty}L^n(\mu)$, but this intersection may be strictly larger than $L^\infty(\mu)$. So I wonder if the greater freedom that we have with $g$ yields a positive answer to the question.

Answer 1

Yes, the answer to your question is positive. For any function $f\in L_1$, such that $f \not\in L_\infty$, we will explicitly find $g\in L_1$, such that $fg \not\in L_1$.

Let us take arbitrary non-negative $f\in L_1, f\not\in L_\infty$. Let $A_k := f^{-1}([k^2, \infty))$, and define $$ g := \sum_k \frac{1}{k^2} \mathbf{1}_{A_k}/\mu(A_k). $$

Clearly $\|g\|_1 \leq \sum k^{-2} < \infty$, yet $\int fg \,\mathrm{d}\mu = \sum k^{-2} \frac{1}{\mu(A_k)} \int_{A_k} f \,\mathrm{d} \mu \geq \sum k^{-2} \cdot k^2 = \infty$.

Answer 2

I think this also a direct consequence of the uniform boundedness principle (see https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Corollaries):

$T_n \colon g \mapsto \int \, f g \mathbb{1}_{\vert f \vert \leq n} \,\mathrm{d}\mu$ is a bounded linear operator $L^1(\mu) \to \mathbb{R}$ and we have the pointwise convergence $T_n(g) \to T(g) = \int \, f g \, \mathrm{d}\mu$ as $n \to \infty$ for all $g \in L^1(\mu)$ by the assumption $\int \, \vert f g \vert \, \mathrm{d}\mu < \infty$ and the dominated convergence theorem. Therefore $T$ is also a bounded operator, i.e. $f \in L^{\infty}(\mu)$.

Thanks to @MaoWao and @DavidGao for this hopefully now correct further remark: The statement "For all $\mathcal{F}$-measurable $f \colon \Omega \to \mathbb{C}$, $\int|fg|\,{\rm{d}}\mu<\infty$ for all $g \in L^1(\mu)$ implies $f \in L^{\infty}(\mu)$" is equivalent to the negation of "There is a sequence of disjoint $\mathcal{F}$-measurable sets $(A_n)_n$ such that for any $n$ it is $\mu(A_n) = +\infty$ and there is no $\mathcal{F}$-measurable subset $B \subseteq A_n$ with $0 < \mu(B) < \infty$" (+) for any measure space $(\Omega, \mathcal{F}, \mu)$.

For one direction, assume that (+) does not hold and that $\int|fg|\,{\rm{d}}\mu<\infty$ for all $g \in L^1(\mu)$. Then there is a (finite) $C > 0$ such that $\Vert f g \Vert_{L^1(\mu)} \leq C \Vert g \Vert_{L^1(\mu)}$ for all $g \in L^1(\mu)$ by the proof above. Suppose $g \notin L^{\infty}(\mu)$. Let $A_n := \{\vert f \vert \in [n, n+1)\}$. Then by assumption there is a $N \in \mathbb{N}$ such that for every $n \geq N$ there is a measurable $B_n \subseteq A_n$ with $0 < \mu(B_n) < +\infty$. Thus $n \mu(B_n) \leq \Vert f \mathbb{1}_{B_n} \Vert_{L^1(\mu)} \leq C \mu(B_n)$ for all $n \geq N$, a contradiction.

For the other direction, assume (+). Let $f = \sum_{n} n \mathbb{1}_{A_n}$ for the sets $A_n$ from (+). Then $\int|fg|\,{\rm{d}}\mu = 0 < \infty$ for all $g \in L^1(\mu)$, but $f \notin L^{\infty}(\mu)$.