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I just could not convince myself about what $\chi_{\cup_{A_n}}$ is worth, if $(A_n)$ is a sequence of pairwise disjoint sets? Is it equal to the series $\sum_{n=1}^{\infty}\chi_{A_n}$? We need to investigate the convergence of the series but how? Many thanks.

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Clearly, $\chi_{\bigcup_{n=1}^{\infty}A_n}(x)$ can assume only two values (it's a characteristic function after all): $1$ or $0$ according as $x\in \bigcup_{n=1}^{\infty}A_n$ or $x\notin\bigcup_{n=1}^{\infty}A_n$.

If $x\in \bigcup_{n=1}^{\infty}A_n$, then $x\in A_m$ for exactly one $m\in\mathbb N$ (given that the union is disjoint), so that $\sum_{n=1}^{\infty}\chi_{A_n}(x)=\chi_{A_m}(x)=1$ (and the series converges because its members vanish for $n>m$). On the other hand, if $x\notin \bigcup_{n=1}^{\infty}A_n$, then $x\in A_m$ for no $m\in\mathbb N$, which implies that $\sum_{n=1}^{\infty}\chi_{A_n}(x)=0$ (and the series converges, since all of its members vanish).

Conclusion: $\chi_{\bigcup_{n=1}^{\infty}A_n}=\sum_{n=1}^{\infty}\chi_{A_n}$, indeed.

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  • $\begingroup$ Thanks a lot triple_sec. I had this idea and thanks for your detailed solution. I posted this as regards to some spectral measure in Rudin...I think I must have misunderstood Rudin's argument as the two contexts are little different. Thanks a lot again. $\endgroup$
    – Math
    May 25 '15 at 8:33
  • $\begingroup$ @Math You're most welcome! $\endgroup$
    – triple_sec
    May 25 '15 at 8:57

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