If $\mu$ is a finite measure and $ν$ is a signed measure with $|\nu|\le C\mu$, are we able to show $\left|\frac{{\rm d}ν}{{\rm d}\mu}\right|\le C$?

Question

Let

• $(\Omega,\mathcal A,\mu)$ be a finite measure space
• $\nu$ be a signed measure on $(\Omega,\mathcal A)$ with $$|\nu(A)|\le C\mu(A)\;\;\;\text{for all }A\in\mathcal A\tag1$$ for some $C\ge0$

Note that $(1)$ implies $\nu\ll\mu$ and hence $$\nu(A)=\int_Af\:{\rm d}\mu\;\;\;\text{for all }A\in\mathcal A\tag2$$ for some $f\in L^1(\mu)$.

I want to show that $$|f|\le C\;\;\;\mu\text{-almost everywhere}\tag3\;.$$ How can we do that?

Answer 1

Hint: For any simple function $g$ with $\|g\|_{L^{1}}\leq 1$, one shows that \begin{align*} \left|\int fgd\mu\right|\leq C, \end{align*} then by letting $f_{n}=f\chi_{E_{n}}$, $E_{n}=\{|x|\leq n, |f(x)|\leq n\}$, one shows that $\|f_{n}\|_{\infty}\leq C$ and $\|f\|_{\infty}=\lim_{n}\|f_{n}\|_{\infty}\leq C$.

Here the converse Holder's inequality/theorem is used.