Background: Let $X$ be a Polish space, that is, a separable completely metrizable topological space. Assume that $F$ is a closed subset of $X.$ Recall that a function $f:X\to\mathbb{R}$ is called Baire-1 if it is a pointwise limit of a sequence of continuous functions. For any $\varepsilon>0$ and $x\in X,$ recall that oscillation of $f$ is defined by $$\omega(f,x,F) = \inf_{\delta>0}\sup_{y,z\in B(x,\delta)\cap F}|f(y)-f(z)|$$ where $B(x,\delta) = \{u\in X: d(u,x)<\delta\}.$ Denote $$D^0(f,\varepsilon,F) = F,$$ $$D^1(f,\varepsilon,F) = \{x\in F: \omega(f,x,F)\geq \varepsilon \}.$$ Clearly $D^1(f,\varepsilon,F)$ is a close subset of $X.$
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Denny and his co-authors proved the following lemma.
Lemma 2.2: Suppose that $F$ is a closed subset of $X$ and that $f$ is a Baire class $1$ function on $X.$ For any $\varepsilon>0,$ there exists a continuous function $g:F\setminus D^1(f,\varepsilon,F)\to\mathbb{R}$ such that $|f(x)-g(x)|<\varepsilon$ for all $x\in F\setminus D^1(f,\varepsilon,F).$
My question is, can we make the bound $|f(x)-g(x)|$ tight? More precisely,
Question: Suppose that $F$ is a closed subset of $X$ and that $f$ is a Baire class $1$ function on $X.$ For any $\varepsilon>0,$ does there exist a continuous function $g:F\setminus D^1(f,\varepsilon,F)\to\mathbb{R}$ such that $f_{|F}=g$ on $x\in F\setminus D^1(f,\varepsilon,F)?$
Another interpretation of my question is that given any Baire-1 function, is it true that the restriction of $f$ to $F\setminus D^1(f,\varepsilon,F)$ continuous for all closed subsets?