I just learn measure theory and I confused about convergence everywhere and convergence almost everywhere.

Let $f_n: \mathbb{R} \to \mathbb{R}$ for all $n$. A sequence $\{f_n\}$ is said converges to $f: \mathbb{R} \to \mathbb{R}$ almost everywhere if $\mu \left({ \left\{{x \in \mathbb{R} : f_n \left({x}\right) \text{ does not converge to } f \left({x}\right) }\right\} }\right) = 0$.

Since an empty set has measure zero, can I conclude that every convergence everywhere is convergence almost everywhere?

  • $\begingroup$ The empty set has measure zero, but there are other sets that have measure zero, like any finite or countably infinite set of real numbers. So for example the function $f_n(x) = \frac{1}{1 + (nx)^2}$ converges to the function $f(x)=0$ almost everywhere (the only exception is at $x=0$). It converges to the following function $g(x)$ everywhere: $$ g(x) = \left\{ \begin{array}{ll} 0&\mbox{ if $x \neq 0$} \\ 1 & \mbox{ if $x=0$} \end{array} \right.$$ $\endgroup$ – Michael Mar 21 '17 at 8:36
  • $\begingroup$ @Michael Yes, I know that the sequence $\{f_n\}$ converges to $f$ a.e. and converges to $g$ everywhere. My question is: if a sequence $\{h_n\}$ converges to $h$ everywhere then can we say it converges to $h$ a.e. ? For example $h_n = h$ is a constant function. Let $A = \{x: h_n(x) \text{ does not converges to } h\}$. Evidently $\mu(A) =0$ so $\{h_n\}$ converges to $h$ a.e. $\endgroup$ – aduh Mar 21 '17 at 9:07
  • $\begingroup$ Yes. Convergence "surely" or "everywhere" implies converges a.e. Convergence everywhere is just a special case of convergence a.e (since the empty set has measure 0). $\endgroup$ – Michael Mar 21 '17 at 20:51

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