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Assume E has finite measure and let $\{f_n\}$ be a sequence of measurable functions on E that converges pointwise to $f$.

Define $E_n = \{ x \in E \ \|\ |f(x) - f_k(x)| < \eta \quad \forall k >=n \}$.

Then the book claims that $E=\cup E_n$ since $f_n$ converges pointwise to $f$. Can anyone explain what this line has to do with pointwise convergence and why does $E=\cup E_n$ hold?

Note: This is from page 65 of Royden's Real Analysis.

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  • $\begingroup$ Pointwise convergence means that for all $x\in E$, $f_n(x)\to f(x)$ as $n\to\infty$. Let $x\in E$, and choose $\eta>0$. What does the definition of convergence tell you about $|f(x)-f_n(x)|$ ? $\endgroup$ – zuggg May 11 '16 at 23:08
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I assume $\eta$ is a fixed positive number. Fix any $x\in E$. Pointwise convergence implies that $f_n(x)\rightarrow f(x)$ as $n\rightarrow\infty$. Thus there exists $m$ such that $|f(x)-f_k(x)|<\eta$ for $k\geq m$, so $x\in E_m\subseteq\cup_n E_n$ (note that $m$ may depend on $x$). Since $x$ was arbitrary, $E\subseteq\cup_n E_n$.

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  • $\begingroup$ Don't you have to prove the other direction? $\endgroup$ – user1559897 May 11 '16 at 23:17
  • $\begingroup$ The other direction is clear from the definition of $E_n$. $\endgroup$ – stewbasic May 11 '16 at 23:19
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If $x \in E$ and $\eta > 0$, find $n$ corresponding to $\eta$ from the convergence of $f_i$ at $x \in E$. Does $x \in E_n?$

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