A function of two variables continues at each variable doesn't have to be continuous but it is pointwise limit of a sequence of continuous functions. In all articles and books that I search for a proof of this statement there is no proof available all saying it is well known that... (with a reference to Baire dissertation more than a century ago). Does some one know any source /book for easy undergraduate based proof?
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Is this function $\mathbb{R}^2 \to \mathbb{R}$? ... or more general? – Brian Moehring Apr 24 '21 at 18:01
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2https://math.stackexchange.com/questions/3019208/separately-continuous-functions-f-mathbbr2-rightarrow-mathbbr-but-now – Alessandro Codenotti Apr 24 '21 at 18:11
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1@AlessandroCodenotti, It is well-known that in this case the set of points of discontinuity is of the first category. like any other places! – Apr 24 '21 at 18:44
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@LiebeGreen That part is not needed to show that a function separately continuous is a limit of continuous functions, it is only needed to show that a separately continuous function is continuous on a big set, but that was not part of your question. In any case I wrote down a short proof of that result which is claimed but not proved in the other answer. – Alessandro Codenotti Apr 24 '21 at 23:35
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To see that any function $\Bbb R^2\to\Bbb R$ separately continuous is of Baire class one (pointwise limit of continuous functions) see the argument given in the answer to this question.
They are assuming the well known fact that every function of Baire class one is continuous on a comeagre set without proof (note that it isn't needed to answer your question about writing a separately continuous function as a limit of continuous ones), but that is the easy part:
Theorem: let $X,Y$ be metrizable spaces with $Y$ separable and let $f\colon X\to Y$ be of Baire class one. Then $f$ is continuous on a comeagre $G_\delta$ set.
Proof: Fix a countable open basis $\{V_n\}$ of $Y$ and note that $f$ is discontinuous at $x$ if an only if there is $n$ such that $x\in f^{-1}(V_n)\setminus\mathrm{int}(f^{-1}(V_n))$. In other words the set of points of discontinuity of $f$ is $$\bigcup_n f^{-1}(V_n)\setminus\mathrm{int}(f^{-1}(V_n)).$$ Now $f^{-1}(V_n)$ is $F_\sigma$ and so is $f^{-1}(V_n)\setminus\mathrm{int}(f^{-1}(V_n))$, so write it as $\bigcup_k F_{n,k}$ with $F_{n,k}$ closed for all $k$. Clearly each $F_{n,k}$ has empty interior, and hence the set of points of discontinuity of $f$ can be written as $\bigcup_n\bigcup_k F_{n,k}$, which is a countable union of closed nowhere dense sets, as we wanted to show.
Alessandro Codenotti
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