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Is it true that for any function of the first Baire class $f:X\to\mathbb R$ on the Cantor cube $X=2^\omega$ there is a continuous function $g:X\to[0,1]$ such that the image $(f+g)(X)$ is disjoint with the set $\mathbb Z$ of integers?

We recall that a function $f:X\to\mathbb R$ is of the first Baire class if it is a pointwise limit of a sequence of continuous functions. By a classical theorem of Baire, a function $f:X\to\mathbb R$ on a Polish space $X$ is of the first Baire class if and only if for any non-empty closed subset $A\subset X$ the restriction $f|A$ has a point of continuity. It is also well-known that each separately continuous function $f:X\times Y\to\mathbb R$ on the product of two Polish spaces is of the first Baire class.

That is why the affirmative answer to the above problem will imply the affirmative answer to the problem Uniform approximation of separately continuous functions on zero-dimensional spaces .

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After some thinking I realized that the answer to this question is negative. A counterexample can be constructed by a standard diagonal method of killing all possible candidatures.

We shall construct a function $f:X\to\mathbb R$ of the first Baire class on the Cantor cube $X=\{0,1\}^\omega$ such that for any continuous function $g:X\to\mathbb R$ the function $f+g$ contains zero in its image.

Observe that the Banach space $C(X)$ of all continuous real-valued functions on $X$ is Polish and hence is the continuous image of the space $\omega^\omega$. Since $\omega^\omega$ is the image of the Cantor cube $X$ under a function of the first Baire class, there exists a surjective function $\xi:X\to C(X)$ of the first Baire class. Now define a function $f:X\to \mathbb R$ by $f(x)=-\xi(x)(x)$ and observe that it is of the first Baire class. For every continuous function $g\in C(X)$ we can find $x\in X$ with $\xi(x)=g$ and conclude that $f(x)+g(x)=-\xi(x)(x)+g(x)=-g(x)+g(x)=0$.

On the other hand, by the proof of the Lebesgue-Hausdorff-Banach Theorem 24.10 in [A.Kechris, Classical Descriptive Set Theory, Springer, 1995], for each bounded function $f:X\to \mathbb R$ of the first Baire class and each $\varepsilon>0$ there is a function $g:X\to\mathbb R$ of the first Baire class such that $\|f-g\|<\varepsilon$ and $g(X)$ is finite.

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