"Almost uniformly convergent" Implies "Uniformly convergent almost everywhere" . Is there something wrong?

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 $$B_{\frac1n}^c=\left[0,1-\frac1n\right]$$.

• 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? Commented Dec 28, 2018 at 14:41
• @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. Commented Dec 28, 2018 at 15:12
• 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 $\langle 0,\infty\rangle$. Commented Dec 28, 2018 at 15:14
• It makes sense. $x$ and $\epsilon$ are both arbitrary in uniformly convergence. Thank you very much. Commented Dec 29, 2018 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 $$\displaystyle \bigcup_n B_{\frac 1n}^c$$.

• 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. Commented Dec 28, 2018 at 14:16
• 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 $f_n(x) \to f(x)$. Commented Dec 28, 2018 at 14:18
• 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? Commented Dec 28, 2018 at 14:27
• 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! Commented Dec 29, 2018 at 3:08