So this question is Chapter 8, Q2, from the book by Wheeden and Zygmund which is related to Hölder's inequality. Let $1\leq p\leq \infty$. A function $f$ is said to be in $L^p(E)$ if $f$ is measurable, and if $\int_E |f|^p$ is finite for $p<\infty$, or if $f$ is essentially bounded for $p=\infty$.
I need to show that given a Lebesgue measurable function $f$ on $E\subset\mathbb R^n$, if $fg\in L^1(E)$ for every Lebesgue measurable $g$ satisfying $g\in L^1(E)$, then $f\in L^\infty(E)$. I notice this question is similar to This question here, but my assumption is a bit weaker.
My attempt is to prove the contrapositive: Assume $\|f\|_\infty=\infty$, then construct a $g\in L^1(E)$ such that $\int_E fg=+\infty$. By using the converse of Hölder's inequality we know that the assumption leads to $$\sup_{\|g\|_1\leq 1}\int_E fg=\infty$$ which means there is a sequence $\{g_n\}\in L^1(E)$ such that $$\lim_{n\to\infty} \int_E fg_n=\infty.$$
This seems to very closed to the result I wanted, but I was hoping a single function $g\in L^1(E)$ can be constructed such that $\int_E fg=+\infty$. I know the unit ball in the Banach space $L^1(E)$ is not compact, generally. Thus it is no guarantee that $\{g_n\}$ has a convergent subsequence, e.t.c. The other attempt was to use Fatou's lemma on the sequence $\{g_n\}$, maybe define $g(x)=\liminf_n g_n(x)$? But I'm not so familiar with the Fatou's lemma, and have no idea is it true that $g\in L^1(E)$ and $\int_E fg$ will be infinite.
How could one construct a $g$ such that $\int_E fg=\infty$? A direct proof is also appreciated.
Let $A_k:=\{x\in\mathbb R^n, \lvert f(x)\rvert\geqslant k\}$. If we assume that $\lVert f\rVert_\infty=\infty$, then $\lambda(A_k)$ is positive for each $k$. But it is possible that $\lambda(A_k)$ is infinite. However, consider $B_k:= A_k\cap B(0, N)$ for $N$ large enough (with $B(0,N)=\{x\in\mathbb R^n,\lvert x\rvert <N\}$) so that $\lambda(B_k)>0$.
Let $(c_k)_{k\geqslant 1}$ be a sequence of positive real numbers such that $\sum_k c_k$ is finite but $\sum_k kc_k$ is infinite and let $$ g=\sum_{k\geqslant 1}a_k\operatorname{sgn}(f)\frac{\mathbf{1}(B_k)}{\lambda(B_k)}, $$ where $\operatorname{sgn}(f)(x)=1$ if $f(x)>0$, $-1$ if $f(x)<0$ and $0$ if $f(x)=0$.
Then $g\in L^1$ but $$fg\geqslant \sum_{k\geqslant 1}a_k k\frac{\mathbf{1}(B_k)}{\lambda(B_k)},$$ which is not integrable.