$fg$ integrable for all $f\in L^1$ iff $g\in L^\infty.$

Question

I have been trying to solve this problem for a week and couldn't come up with a solution.

Let $g $ be a Lebesgue measurable function defined on $\mathbb R$. Prove that the following are equivalent:

a) The function $g$ is bounded (i.e., $g\in L^\infty(\mathbb R)$.

b) For every function $f \in L^1(\mathbb R)$ we have $f(x) g(x)\in L^1(\mathbb R)$

That a) $\Rightarrow$ b) is quite straightforward, if $g\in L^\infty(\mathbb R)$ then there exists a constant $C = inf \{ c:\mu(\{x:|g(x)|\geq c\})=0\} $, therefore, for every function $f \in L^1(\mathbb R)$ we have $$\int| f(x) g(x)|d\mu \leq \int |f(x)|Cd\mu < \infty $$ Which means $f(x) g(x)\in L^1(\mathbb R)$.

For the converse, I thought about negating and proving by contradiction. This is, if $g\notin L^\infty(\mathbb R)$, then there exists a function $f \in L^1(\mathbb R)$, but $f(x) g(x)\notin L^1(\mathbb R)$

So far, my idea is that we know that for every constant $k$, $\mu(\{x:|g(x)|\geq k\}) > 0 $, because if it is $0$ then we already have a contradiction, as that means $g\in L^\infty(\mathbb R)$.

I thought about creating characteristic functions using those decreasing sets, as for each $k$ we are able to obtain a set. Moreover, if $k$ increases then the size of the set decreases. But I am unable to move forward from here.

Any hints or solutions will be appreciated.

Answer 1

Suppose $fg\in L^1$ for all $f\in L^1$ but $g\notin L^\infty.$ (We are on $\mathbb R$ with Lebesgue measure $\mu.$) Then $|g|^{-1}([n,n+1))$ are pairwise disjoint and measurable, and have positive measure for infinitely many $n,$ say along the subsequence $n_1<n_2 <\cdots.$ Choose $E_k \subset |g|^{-1}([n_k,n_k+1))$ such that $0<\mu(E_k)<\infty.$ Now define

$$f=\text {sgn } g\cdot\sum_{k=1}^{\infty}\frac{1}{k^2}\frac{\chi_{E_k}}{\mu(E_k)}.$$

Then $f\in L^1,$ yet

$$\int_{\mathbb R} fg\,d\mu= \sum_{k=1}^{\infty}\frac{1}{k^2}\frac{1}{\mu(E_k)}\int_{E_k}|g|\, d\mu$$ $$ \ge \sum_{k=1}^{\infty}\frac{1}{k^2}\cdot n_k \ge \sum_{k=1}^{\infty}\frac{1}{k^2}\cdot k = \infty,$$

contradiction.

Answer 2

One way of proving this is to use Uniform Boundedness Principle. Define $T_n: L^{1} \to \mathbb R$ by $T_n(f)=\int_{\{|g|\leq n\}} fg$. Then $T_n$ is a a bounded linear functional and $(T_nf)_n$ is bounded for each fixed $f$ (because $\int |fg| <\infty$). Hence, by Uniform Boundedness Principle, $sup_n ||T||_n <\infty$. Now let $E=\{x:|g(x)| >N\}$ and assume that $\mu(E)>0$. Consider $T_n(\frac g {|g|} \frac {I_E} {\mu(E)})$. Note that $\frac g {|g|} \frac {I_E} {\mu(E)}$ has norm $1$. Hence $\|T_n\| \geq T_n(\frac g {|g|} \frac {I_E} {\mu(E)})=\frac 1 {\mu(E)}\int_E |g| d\mu>N$. This cannot hold for $N$ sufficiently large (because $sup_n ||T||_n <\infty$). Hence $\mu (E)$ must be $0$ and $|g| \leq N$ if $N$ is sufficiently large.