If we have in certain norm


1). $g\_j(x) \rightarrow h(x), j\rightarrow\infty$ and 


2). $ f\_{ij}(x)\rightarrow g\_j(x), i\rightarrow \infty, \forall j$ , 


then we can choose a subsequence $\{f\_{ij\_{(i)}}, i=1,2,...\}$ from $\{f\_{ij}, i=1,2,...;j=1,2,...\}$ such that $f\_{ij\_{(i)}}\rightarrow h, i\rightarrow\infty$, in that norm sense. And it's just a trivial matter of triangle inequality. (I leave some freedom for function definition, so that more examples or counterexamples can be taken into account.)


Now my question is: 


What if the convergence is in pointwise sense? Can we still do that? namely, 
if $g\_j(x) \rightarrow h(x), j\rightarrow\infty$ pointwise,and if $ f\_{ij}(x)\rightarrow g\_j(x), i\rightarrow \infty, \forall j$ pointwise, can we find a subsequence $f\_{ij\_{(i)}}$ from $f\_{ij}$ such that $f\_{ij\_{(i)}}\rightarrow h, i\rightarrow\infty$ pointwise?


I constructed a counterexample which seems to be complicated. I wonder if there is any easy answer. Thanks.



approximating functions pointwise
Let $h$ be the indicator function of the rationals. This is of Baire class 2, i.e. it is not the pointwise limit of a sequence of continuous functions, but it is the pointwise limit of a sequence of pointwise limits of continuous functions.
For example, let $r\_1, r\_2, \ldots$ be an enumeration of the rationals, 
$g\_n(x)$ the indicator function of $S\_n = \{r\_1, r\_2, \ldots, r\_n\}$, and
$f\_{ij}(x) = \exp(-i \;\text{dist}(x, S\_j))$. 


