I'm trying to compile a list of counterexamples for convergence implications (or rather, the lack of). I have an incomplete list and I hope to get it all together in one piece.
I'm currently working with the 5 different notions of convergence: uniformly, pointwise, a.e., $L^1$ and in measure.
I know that we have the following implications:
uniformly $\implies$ pointwise $\implies$ a.e.
uniformly $\implies$ in measure
and finally, $L^1 \implies$ in measure.
That means I should have counterexamples for the following:
pointwise $\not\Rightarrow$ uniformly: $f_n(x)=x^n$ on $[0,1]$
pointwise $\not\Rightarrow$ in measure: $f_n(x)=\chi_{[n,\infty)}$, or $f_n(x)=\chi_{[n,n+1]}$
pointwise $\not\Rightarrow L^1$: $f_n = 1$ when $x\in \mathbb{Q}$ and $f_n(x) = 1/n$ otherwise
a.e. $\not\Rightarrow$ pointwise: $\{r_n\}_{n\in\mathbb N}$ is an enumeration of $\mathbb Q$, $f_n(x)=(x-r_n)^{-\frac{1}{2}}$ if $r_n<x\leq r_n+1$ and $f_n(x)=0$ otherwise
a.e. $\not\Rightarrow$ uniformly: $f_n(x)=x^n\chi_{[0,1]}$
a.e. $\not\Rightarrow L^1$: $f_n(x)=\chi_{[n,n+1]}$
a.e. $\not\Rightarrow$ in measure: $f_n(x)=\chi_{[n,n+1]}$
in measure $\not\Rightarrow L^1$: $f_n(x)=n\chi_{[0,\frac{1}n]}$
in measure $\not\Rightarrow$ uniformly: $f_n=n\chi_{[0,\frac{1}{n^2}]}$
in measure $\not\Rightarrow$ pointwise:
in measure $\not\Rightarrow$ a.e.: $f_1=\chi_{[0,1]},f_2=\chi_{[0,\frac{1}{2}]},f_3=\chi_{[\frac{1}{2},1]},f_4=\chi_{[0,\frac{1}{4}]},\cdots$
$L^1\not\Rightarrow$ uniformly: $f_n=n\chi_{[0,\frac{1}{n^2}]}$
$L^1\not\Rightarrow$ pointwise: $f_n = \chi_{[0,1/n]}$
$L^1\not\Rightarrow$ a.e.: $f_n=\chi_{I_n},I_n=[j2^{-k},(j+1)2^{-k}]$ for $n=2^k+j$
uniformly $\not\Rightarrow L^1$: $f_n(x)=\frac{1}n\chi_{[0,n]}$
As people provide solutions, I will try to update this accordingly. 🙂
Best Answer
a.e. $\nRightarrow $ uniformly : $f_n(x)=x^n\chi_{[0,1]}$. It converges a.e. to 0 but not uniformly.
a.e.$\nRightarrow$ $L^1$: $f_n(x)=\chi_{[n,n+1]}$ converges a.e. to 0, but the integrals are all 1.
a.e.$\nRightarrow$ in measure: use the above example: $\mu(\{f_n\geq\varepsilon\})=\mu(\{f_n=1\})=1\nrightarrow 0$
in measure $\nRightarrow$ uniformly: $f_n=n\chi_{[0,\frac{1}{n^2}]}$
I will update the list