bounded function in $\mathbb{R}^d$

Question

There is a classic problem in Analysis stating that for integrable $f$ and $g$ bounded and measurable then their product is integrable. Now I am trying to show if $g$ is measurable on $\mathbb{R}^d$ such that its product with $f$ is integrable on $\mathbb{R}^d \ \forall f$ integrable, then $g$ is bounded outside a set of measure $0$.

Answer 1

This problem is solved by looking at a specific sequence of sets, namely the sets $A_n = \{x \mid n \leq g < n+1\}$.

We work by contrapositive. If $g$ is unbounded, we want to find a function $f \in L^1$ such that $fg \notin L^1$. Since $g$ is not bounded outside a set of measure 0, there are infinitely many sets $A_n = \{x \mid n \leq g < n+1\}$ such that $\mu(A_n) > 0$. Without loss of generality, assume this happens for a positive sequence of integers $n_k$ (consider $-g$ otherwise), which may be taken strictly increasing.

We have a strictly increasing sequence $n_k$ such that $\mu(A_{n_k})>0$. Since it is strictly increasing, $n_k \geq k$. Let us try to find the desired $f$ as $$ f = \sum_{k=1}^\infty c_k 1_{A_{n_k}}$$ for a well chosen sequence of positive real numbers $c_k$. We want $\int f < \infty$ and $ \int fg = \infty$.

$$ \int f d\mu = \sum c_k \mu(A_{n_k}), $$ $$ \int fg d\mu = \sum c_k \int_{A_{n_k}} g d\mu \geq \sum c_k \mu(A_{n_k}) \ n_k $$ since $g \geq n_k$ on $A_{n_k}$. We want to find a sequence $c_k$ such that the first sum is finite and the second sum is infinite.

We can take $$c_k = {1\over \mu(A_{n_k})} {1 \over k \, n_k}.$$ The first sum is equal to $\sum {1\over k \, n_k} \leq \sum {1 \over k^2} < \infty$ since $n_k \geq k$. The second one is equal to $\sum {1\over k} = \infty$.