# $fg\in L^1$ for every $g\in L^1$ prove $f\in L^{\infty}$

Let $(X,\mathcal{A}, \mu)$ be an arbitrary measure space. Let $f$ be an extended complex-valued $\mathcal{A}-$measurable function on $X$ such that $|f|<\infty$ $\mu$-a.e. on $X$. Suppose that $fg\in L^1(X,\mathcal{A}, \mu)$ for every $g\in L^1(X,\mathcal{A}, \mu)$. Show that $f\in L^{\infty}(X,\mathcal{A}, \mu)$.

1. Can anyone verify my answer?
2. Does anyone know a better elementary approach? (It’ll be great if similar approach can be generalized to the case of $L^p$ and $L^q$)

Related question: On $\sigma-$ finite space $fg\in L^1$ for every $g\in L^q$ prove $f\in L^p$

For any given $f\notin L^{\infty}$, define $E_n=\{x\in X|n-1<f(x)\leq n\}$, there is a subsequence $E_{n_1}$,…$E_{n_j}$ such that $\mu(E_{n_j})>0$ for each $j$. Define $\displaystyle g=\sum_{j=1}^{\infty}\frac{1}{j^2\mu(E_{n_j})}\mathbb{1}_{E_{n_j}}$, we have $\displaystyle \int_{X}|g|d\mu=\sum_{j=1}^{\infty}\frac{1}{j^2}<\infty$ so $g\in L^1$. However $$\int_{X}|fg|d\mu=\int_{X} \sum_{j=1}^{\infty} \frac{|f|\mathbb{1}_{E_{n_j}}}{j^2\mu(E_{n_j})}$$
because $f$ is finite almost everywhere and the last expression
$$\int_{X} \sum_{j=1}^{\infty} \frac{|f|\mathbb{1}_{E_{n_j}}}{j^2\mu(E_{n_j})}>\int_{X} \sum_{j=1}^{\infty} \frac{(n_j-1)\mathbb{1}_{E_{n_j}}}{j^2\mu(E_{n_j})}>\int_{X} \sum_{j=1}^{\infty} \frac{(j-1)\mathbb{1}_{E_{n_j}}}{j^2\mu(E_{n_j})}=\sum_{j=1}^{\infty}\frac{j-1}{j^2}=\infty$$ by limit comparison with harmonic series. Therefore $fg\notin L^1$

#### Solutions Collecting From Web of "$fg\in L^1$ for every $g\in L^1$ prove $f\in L^{\infty}$"

If I’m not mistaken this result is not true on an arbitrary measure space. For example consider the measure $\mu$ defined by
\begin{align*}
\mu(A) = \begin{cases}
0 & \mbox{if } A = \emptyset \\
\infty & \mbox{otherwise}
\end{cases}
\end{align*}
Then $g \in L^1(\mu)$ iff $g = 0$. As a result $fg = 0 \in L^1(\mu)$ for all measurable functions $f$. But $f \in L^\infty(\mu)$ iff $f$ is bounded so now we just pick measurable, finite $f$ which is not bounded for a counterexample.

I believe the problem with your proof is that, as pointed out in the comments, we could have $\mu(E_n) = \infty$ and worse $E_n$ could have no measurable subsets of positive, finite measure, as in the example above. The right condition on $\mu$ will be that $\mu$ is semifinite so that $E_n$ has a subset of positive, finite measure and working with this subset should make your proof go through.