I am reading Kechris's Classical Descriptive Set Theory. Kechris prove the following theorem at page $16.$
Theorem $3.8$ (Kuratowski) Let $X$ be metrizable, $Y$ be completely metrizable, $A\subseteq X,$ and $f:A\to Y$ be a continuous function. Then there exist a $G_\delta$ set $G$ with $A\subseteq G\subseteq \overline{A}$ and a continuous extension $g:G\to Y$ of $f.$
The following is a definition of $G_\delta.$
Let $$G = \overline{A}\cap\{x:\text{osc}_f(x)=0\}.$$ This is a $G_\delta$ set and since $f$ is continuous $A,$ $A\subseteq G\subseteq \overline{A}.$
Then the proof proceeds with the standard construction of $g$ on $G.$ (Essentially, for any $x\in\overline{A},$ get a sequence $(x_n)$ converging to $x.$ Then the sequence $(f(x_n))_n$ is Cauchy and thus has a limit in $Y.$ Define $g(x)$ to be the limit.)
The following is my question.
Question: Is $G=A?$ Since $f$ is continuous on $A,$ so $\{x:\text{osc}_f(x)=0\} = A.$ Am I right?