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](https://mathoverflow.net/questions/218554/uniform-approximation-of-separately-continuous-functions-on-zero-dimensional-spa) .



On continuous perturbations of functions of the first Baire class on the Cantor set
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. 


