What should I use in order to get the required function $g \in L^1$?

Question

Let $f \in L^1$ be such that $f$ is not equivalent to any bounded function. Prove there exists a function $g \in L^1$ such that $fg \notin L^1$.

I know that $m( \{x:|f(x)|>M\} )>0$, $\forall M>0$, but I can't see to come up with anything else useful to use for this problem in order to get the required function.

Thank you for any hints as to how may I go about approaching this problem.

Answer 1

One can give a simple constructive proof of existence of such $g$.

Assume we have $f\in L^1(0,1)$ which is not $L^\infty$ (we fix $(0,1)$ for the sake of clarity, this does not impose any restriction on the construction below). For each $n\in \mathbb{N}$ define $$ E_n = \{ x \in (0,1): n\leq |f(x)| \}. $$

Since $f\notin L^\infty$, then $\mu(E_n) >0$ for all $n\in \mathbb{N}$. The aim now is to construct a function $g\in L^1$ such that $fg \notin L^1$. To this end, set $$ (1) \qquad g = \sum_{n=1}^\infty \frac 1n a_n \chi_{E_n}, $$ where $a_n>0$ is a sequence of positive numbers to be fixed in a moment. We thus have $$ (2) \qquad \int|fg| d\mu = \sum_{n=1}^\infty \frac{a_n}{n} \int_{E_n}|f| d\mu \geq \sum_{n=1}^\infty \frac{a_n}{n} n \mu(E_n) = \sum_{n=1}^\infty a_n\mu(E_n). $$

Hence, to complete the construction we need to fix $a_n$ so that

(a) $\sum_{n=1}^\infty \frac{1}{n} a_n \mu (E_n) < \infty$ which is the condtion of $g\in L^1$ (see $(1)$ )

(b) $\sum_{n=1}^\infty a_n \mu (E_n) = \infty$, which amounts to $fg \notin L^1$ (see $(2)$ ).

Thanks to $f\notin L^\infty$, we have $\mu(E_n) >0 $ for all $n$ and hence setting $a_n = \frac{1}{\mu(E_n)} \frac{1}{n^{1/2}}$ we get both (a) and (b) and hence complete the construction of $g$.

Answer 2

I don't think there is a simple measure theoretic proof of this. Completeness of $L^{1}$ is crucial. Here is the standard functional analytic proof: define linear functionals $\Phi_n$ on $L^{1}$ by $\Phi_n (g)=\int I_{|f| \leq n} fg$. These are continuous linear functionals on $L^{1}$ and $\sup_{\{n \geq 1\}} |\Phi_n (g)| <\infty$. Now apply Uniform Boundedness Principle to conclude that $\sup ||\Phi_n|| <\infty$. A standard argument used in the proof of $(L^{1})^{*} =L^{\infty}$ show that $||\Phi_n||=||I_{|f| \leq n} f||_{\infty}$. Hence $f$ is (essentially) bounded.