Let $f$ be a bounded real-valued measurable function. I want to compute the the limit of the integral $$\int f\chi_{A_n}\,\mathrm d\mu$$ as $n$ approaches infinity. Here $\chi_{A_n}$ denotes the indicator of the set $A_n := \{\vert f\vert\geq \sqrt n\}$. If we assume that the measure $\mu$ is finite, $\sup\vert f\vert$ is integrable and dominates $f\chi_{A_n}$. Thus, $$\lim_{n\rightarrow\infty}\int f\chi_{A_n}\,\mathrm d\mu = \int\lim_{n\rightarrow\infty}f\chi_{A_n}\,\mathrm d\mu$$ by dominated convergence. Now since $A_n\supseteq A_{n+1}$ for all $n\in\mathbb N$, it follows that $\lim_{n\rightarrow\infty}\chi_{A_n} = 0$. Hence the value of the integral is zero in the limit.
Is my reasoning correct?