Let $E$ be a measurable set and let $g$ be a function defined on $E$. The product $fg$ belongs to $L^1(E)$ for every function $f \in L^1(E)$

Question

Let $E$ be a measurable set and let $g$ be a function defined on $E$. The product $fg$ belongs to $L^1(E)$ for every function $f \in L^1(E)$ if and only if $g ∈ L^{\infty}(E)$.

Answer 1

Sketch of a home-made proof: I assume we are in a $\sigma$-compact measure space. WLOG, $g\ge 0.$ Suppose $g\notin L^\infty.$ Then for any $a>0,$ there exists $b>a$ such that $0<\mu(\{a<g<b\})< \infty,$ here using $\sigma$-compactness. Using this repeatedly, we can choose $a_k > 2^k$ such that $0<\mu(\{a_k<g<a_{k+1}\})< \infty.$ Let $E_k = \{a_k<g<a_{k+1}\}.$ Set

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

Then $f\in L^1.$ But

$$\int fg = \int \sum_{k=1}^{\infty} g\frac{1}{k^2\mu(E_k)}\chi_{E_k} \ge \sum_{k=1}^{\infty} \frac{2^k}{k^2} = \infty.$$

That's a contradiction, giving us the result.