# Prove that { $\langle M \rangle$ : $M$ is a TM and $L(M)$ is decidable} is undecidable

So I want to prove that $$\big\{\langle M \rangle : \text{ M is a TM and } L(M) \text{ is decidable} \big\}$$ is undecidable.
To do so I want to reduce it from$$\ \overline{A_{TM}}$$ with a function which looks like this :
On input $$\langle M,w \rangle$$ run $$M$$ on $$w$$, if $$M$$ accepts $$w$$ output $$M'$$ where $$M'$$ should be a TM for an undecidable language.
Furthermore the function loops if $$M$$ does not accept $$w$$ such that if $$M$$ reject $$w$$, $$M'$$ loops on every input.
Hence if $$\langle M,w \rangle \in \overline{A_{TM}}$$, $$M$$ rejects $$w$$ and $$L(M') = \emptyset$$ which is decidable.
My problem is when $$\langle M,w \rangle \in A_{TM}$$ and I want to output a TM $$M'$$ such that $$L(M')$$ is undecidable but I don't know how to create such a TM.

Maybe try to use Rice' theorem, instead of reducing from $$\overline{A_{TM}}$$.
We only need a TM that recognizes an undecidable language so we just have to take a TM $$M'$$ that recognizes $$A_{TM}$$ for example and return it when $$M$$ accepts $$w$$.
We have $$L(M') = A_{TM}$$ which is undecidable.