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

Question

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.

Answer 1

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

Answer 2

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