Let $(\Omega,\mathcal{A},\mu)$ be a non-atomic, $\sigma$-finite measure space. If $f \in L^{\infty}(\Omega)$ satisfies that "for every sequence $(\Gamma_n)_{n=1}^{\infty} $ of measurable subsets of $\Omega$ for which $\chi_{\Gamma_n} \rightarrow 0 \ \mu$-a.e. then $\|f \cdot \chi_{\Gamma_n}\|_{\infty} \rightarrow 0$" then $f=0 \ \mu$-a.e.
I read about this result in the book "Interpolation of operators" by Bennett and Sharpley (p.30) but I wasn't able to find or produce a proof. Some help or reference would be appreciated.
The measure $\mu$ is called non-atomic if for every $A \in \mathcal{A}$ with $\mu(A)>0$ there exists a measurable subset $B$ of $A$ such that $0<\mu(B)<\mu(A)$.
[Edit] An attempt has been made to produce a proof in the comments but it is still unclear if that proof really works and if it does (it seems that it doesn't), how non-atomicity is used in order to get that $\|f \cdot \chi_{\Gamma_n}\|_{\infty}>\varepsilon$ .
If $f\neq 0$ on a set of positive measure, then there exists $\epsilon>0$ and a measurable set $\Gamma$ with positive measure such that $|f|\geq \epsilon$ on $\Gamma$. Since $\mu$ is $\sigma$-finite, we can assume that $\Gamma$ has finite measure. By Simpler proof - Non atomic measures, there exists a descending sequence $(\Gamma_n)$ of measurable subsets of $\Gamma$ such that $\mu(\Gamma_n)=2^{-n}$. Then $\chi_{\Gamma_n}\to 0$ a.e. and $\|f\chi_{\Gamma_n}\|_\infty\geq \epsilon$.