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Here is a very intuitive definition of convergence almost anywhere from ProofWiki:

Sequence of function $(f_n)_{n \in \mathbb N}$ is said to converge almost everywhere on $ D$ to $ f$ if and only if $$\mu (\{x \in D \mid f_n(x) \text { does not converge to } f(x)\})=0$$ and we write it as $$f_n \xrightarrow {a.e.} f. $$

Intuitively, it means that the measure of the set, containing all the $x$'s that do not make $f_n$ converge to $ f $, is zero. But here comes my question: How does the adjective "almost everywhere" fit in? Am I correct in saying that the adjective "almost everywhere" means there is a few being left out because the way $\mu $ is defined?

Thank you for your time and effort.

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  • $\begingroup$ As in "for almost every $x$". It formalises this idea.. - Does it converge on the whole real line? - No, it does not, but it converges almost everywhere! $\endgroup$ – Ant Mar 31 '15 at 21:30
  • $\begingroup$ To add to the given answers: one important reason why we only care about certain properties almost everywhere is because, at least with respect to integration, if two functions are equal almost everywhere, then their integrals are equal. So if I have a function $f$, I can change its value on a set of measure $0$ with respect to the measure $\mu$, and the new function will have the same integral as $f$. $\endgroup$ – layman Mar 31 '15 at 21:43
  • $\begingroup$ Thanks to both of you guys! $\endgroup$ – Amanda.M Mar 31 '15 at 21:57
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Am I correct in saying that the adjective "almost everywhere" means there is a few being left out because the way μ is defined?

Yes, exactly that. Take for example the Lebesgue measure on $\mathbb{R}$. Every countable set is a null-set w.r.t the Lebesgue measure, that means if your sequence $f_n$ converges to $f$ on all $x \in \mathbb{R}$ except for countably many $x$, this would still be convergence almost everywhere.

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  • $\begingroup$ Thank you! Now I am an happy camper. $\endgroup$ – Amanda.M Mar 31 '15 at 22:00
  • $\begingroup$ You're welcome, I'm glad I could help. $\endgroup$ – Barkas Mar 31 '15 at 22:22
  • $\begingroup$ The convergence being spoken about here is pointwise, correct? $\endgroup$ – Anant Joshi May 26 '18 at 18:18
  • $\begingroup$ Yes, that is pointwise. $\endgroup$ – Barkas May 29 '18 at 8:20
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To put it another way, there exists a set $E$ with $\mu(E) = 0$ such that $\lim_{n\to \infty} f_n(x) = f(x)$ for all $x\in D\setminus E$. The "almost everywhere" piece comes in as $f_n$ converges pointwise to $f$ eveywhere except possibly on $E$.

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    $\begingroup$ As always, thank you for your extra example! $\endgroup$ – Amanda.M Mar 31 '15 at 22:01
  • $\begingroup$ When it says "converges everywhere" it means that the function "converges pointwise" ? $\endgroup$ – user178403 Apr 14 at 19:35
  • $\begingroup$ @user178403 convergence (almost) everywhere means convergence pointwise (almost) everywhere. $\endgroup$ – kobe Apr 14 at 19:39
  • $\begingroup$ oh ok. Though it's not common to hear 'convergence pointwise everywhere', just 'convergence pointwise'. Or is there a difference among the two ? $\endgroup$ – user178403 Apr 14 at 19:42
  • $\begingroup$ @user178403, no there is no difference. Convergence pointwise is the standard terminology. The word ‘everywhere’ would go along with ‘almost’ as in “convergence almost everywhere.” $\endgroup$ – kobe Apr 14 at 19:47

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