At first I don't know if $Y$ is measurable, we only know that it has a positive outer measure and is dense. So I don't know how to get to a contradiction.

$1=\lim_{h\to 0}\frac{m( (A\cap Y\cap]-h,h[) \cup (A\setminus Y\cap]-h,h[))}{2h}=\lim_{h\to 0}\frac{m( (Y\cap [0,h[) \cup ((A\setminus Y)\cap]-h,h[))}{2h} $.

Then $0\in A\cap Y$, so $1=\lim_{h\to 0}\frac{m(A\cap ]-h,h[)}{2h}$.

Now, my question is why $Y\cap [0,+\infty[$ is not open in $Y$?. Otherwise, there is $A\in\mathcal{T}$, such that $Y\cap [0,+\infty[=A\cap Y$. As $Y$ is dense, then $\mbox{int}_{\mathcal{T}}(A\setminus Y)=\emptyset$.

Thanks @KP Hart. In fact, $\overline{Y\cap ]0,+\infty[}^{Y}=Y\cap \overline{Y\cap ]0,+\infty[}^{\mathcal{T}}=Y\cap \overline{]0,+\infty[}^{\mathcal{T}}=Y\cap [0,+\infty[ $, because $Y$ is dense,

Hi Stefan, do you have any idea to prove that $\omega^{\omega}$ is a homogeneous space?

Thanks for the information professor Alessandro

Thanks for the information Professor Mohammad.

The reference on this question is Lemma 1.0 of the article Products of Baire spaces by Paul Cohen.

Well, in this case I understand that $(P, \tau_{\leq})$ as topological space is Baire, that is, every countable family of open and dense sets in $P$ has dense intersection.

I have one last question, I can prove that $D_ {n}$ is open and dense at $\{q: q\leq p^{\prime} \}$, then $\bigcap _{n\in \omega}D_{n}$ is dense in $\{q: q\leq p^{\prime} \}$, but I need that $G$ intersects $\bigcap _{n\in \omega}D_{n}$, and this will happen if $\bigcap _{n\in \omega}D_{n}$ is dense. How can I get that $\bigcap _{n\in \omega}D_{n}$ is dense in $P$?

I understood, in addition to the proof it is enough that they are dense below $p^{\prime}$, since an open subspace of a Baire remains Baire.