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I started working on a problema on building a sequence of continuous function whose pointwise limit has to be a real valued function f. I mean: $f:R^2 \rightarrow [-\infty; +\infty]$ and I know that the two function fixing the single variables defined as:
h:$t \rightarrow f(x,t)$
g:$t \rightarrow f(t,y)$
are continuous. I'm asked to prove that f is the pointwise limit of a sequence of continuous function (and since it is proved it also measurable).

So I started working with the variables y fixed;imaging that if I consider f for each $(x,y+\frac{1}{2^n})$ the succesion $f_n=f(x,y+\frac{1}{2^n})$.
I can get that this tends to f (using the conitinuity of g) but $f_n$ as defined is not continuous.So I don't know how to solve this problem.May I use some difference between two points as continuous function (where the continuity is given by the functions g and h)? I'm in lack if ideas..

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You may construct $f_n$ as follows.

$f_n(x,y)=(k+1-2^ny)f(x,\frac{k}{2^n})+(2^ny-k)f(x,\frac{k+1}{2^n}),\quad \frac{k}{2^n}\le y<\frac{k+1}{2^n}, k\in\mathbb{Z}.$

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    @Laura: It is not accurate to say "$k$, $n$ are fixed for every $y$", because if $y$ and $n$ are given first, the choice of $k$ depends on $n$ and $y$. Nevertheless, if $k$, $n$ are given first, on the interval $[\frac{k}{2^n},\frac{k+1}{2^n}]$, as you have seen, $f_n$ is a sum of two terms, and both terms are product of two continuous functions, so $f_n$ is continuous.2012-11-21