Inequality on “supremum” of product of functions

measure-theoryself-learning

Let $(A,\mathcal{F},\mu)$ be a measure space and $f$ a measurable real
function. Define $\text{ess} \sup (f)=\inf\{c\in \mathbb{R}:\mu(\mid f\mid>c)=0\}$ and $\text{ess} \inf (f)=\sup\{c\in \mathbb{R}:\mu(\mid f\mid<c)=0\}$.

I want to show that $\text{ess} \sup (fg)\leq \text{ess} \sup(f). \text{ess} \sup(g)$, when both sides are nonnegative.

Can you give me a hint to prove it?

Best Answer

Note that it's enough to show that $\mu(|fg|> \text{ess sup}(f) \cdot \text{ess sup}(g)) = 0$. If $|fg| > \text{ess sup}(f)\cdot \text{ess sup}(g)$, then either $|f|>\text{ess sup}(f)$ or $|g|> \text{ess sup}(g)$ (why is this?).

To complete the proof, you need to show that $\mu(|f| > \text{ess sup}(f)) = 0$ (and the same will follow for $g$). Two facts are important here:

  • $|f|> \text{ess sup}(f)$ if and only if there exists $n\in \mathbb{N}$ such that $|f|> \text{ess sup}(f)+\frac{1}{n}$.
  • By definition of infimum, for every $n$, there exists $c_n < \text{ess sup}(f)+\frac{1}{n}$ such that $\mu(|f|>c_n)=0$.

Use these two facts to conclude that $\mu(|f|>\text{ess sup}(f))=0$.

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