I find a tricky proof shows "almost uniformly convergent" implies "uniformly convergent almost everywhere". I know it is wrong, for there is a counterexample. Can anyone help me why this proof is wrong?

I got the inspiration from the proof that shows "almost uniformly convergent" implies "convergent almost everywhere":

It says $\forall \epsilon, \exists B_\epsilon, \mu(B_\epsilon)<\epsilon$ outside of which $f_n$ converges uniformly to $f$ and prove $B\equiv\bigcap_{n\in\mathbb{N}}B_{\frac{1}{n}}$ is zero-measure. I think in this case, outside $B$, $f_n \to f$ no only converges pointwise (as the proof says) but also converges uniformly (for it is true outside any $B_\frac{1}{n}$). Since $\mu(B)=0$, it implies $f_n \to f$ converges uniformly almost everywhere.


Consider a simple example:

$x^n \to 0$ almost uniformly on $[0,1]$. Indeed, for every $\varepsilon > 0$ we have $x^n \to 0$ uniformly on $B_\varepsilon^c = [0,1-\varepsilon]$ where $B_\varepsilon = [1-\varepsilon, 1]$ has measure $\varepsilon$.

However on $$B^c = \left(\bigcap_{n=1}^\infty B_{\frac1n}\right)^c = \left(\bigcap_{n=1}^\infty \left[1-\frac1n,1\right]\right)^c = \{1\}^c = [0,1\rangle$$ the convergence $x^n \to 0$ is not uniform. It is only uniform on segments $B_{\frac1n}^c=\left[0,1-\frac1n\right]$.

  • $\begingroup$ Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{\frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference? $\endgroup$ – Xuchuang Dec 28 '18 at 14:41
  • $\begingroup$ @Xuchuang On $B_{\frac1n}^c$ we have $f_n \to f$ uniformly, and in particular pointwise. Then for every $x \in B^c$, we have $x \in B_{\frac1n}^c$ for some $n \in \mathbb{N}$ so $f_n(x) \to f(x)$. Since $x$ was arbitrary, we conclude $f_n \to f$ pointwise on $B^c$. We cannot conclude $f_n \to f$ uniformly as the example shows. $\endgroup$ – mechanodroid Dec 28 '18 at 15:12
  • $\begingroup$ It's sort of like how the function $x \mapsto \frac1x$ is uniformly continuous on all segments $[a,b] \subseteq \langle 0, \infty\rangle$ but it is not uniformly continuous on entire $\langle 0,\infty\rangle$. $\endgroup$ – mechanodroid Dec 28 '18 at 15:14
  • $\begingroup$ It makes sense. $x$ and $\epsilon$ are both arbitrary in uniformly convergence. Thank you very much. $\endgroup$ – Xuchuang Dec 29 '18 at 3:02

You have that $\{f_n\}$ converges uniformly on each set $B_{\frac 1n}^c$. There is no reason why it should converge uniformly on $\displaystyle \bigcup_n B_{\frac 1n}^c$.

  • $\begingroup$ Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $\bigcup_n B_{\frac{1}{n}}^c$. Is there a difference between them. $\endgroup$ – Xuchuang Dec 28 '18 at 14:16
  • $\begingroup$ That isn't to hard to see. If a point $x$ belongs to $\bigcup_n B^c_{\frac 1n}$, then there exists (at least one) index $n$ with $x \in B_{\frac 1n}^c$. This means that $f_n(x) \to f(x)$. $\endgroup$ – Umberto P. Dec 28 '18 at 14:18
  • $\begingroup$ But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right? $\endgroup$ – Xuchuang Dec 28 '18 at 14:27
  • $\begingroup$ You are right. The wrong place is when unionizing those $B_{\frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you! $\endgroup$ – Xuchuang Dec 29 '18 at 3:08

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