As seeing the comment on the 2nd answer of this question, mentioning delta function, I tried to write the proof of that part.
Proof
As $T(\mathbf{x})=t\mathbf{1}_{\mathbf{X}}(\mathbf{x}) \Rightarrow \mathbf{X} = \mathbf{x}$, $\mathbf{X}=\mathbf{x}$ is necessary to $\mathbf{T}(\mathbf{x})=t$.
Therefore, $\{\{\omega \mid\mathbf{X}=\mathbf{x}\} \cap \{\omega \mid\mathbf{T}(\mathbf{x})=t\}\}=\{\omega \mid\mathbf{X}=\mathbf{x}\}$.
$\mathbf{1}_{\mathbf{X}}(\mathbf{x})$ is an indicator function, which is defined
$$ {\displaystyle \mathbf {1} _{X}(x):={\begin{cases}1~&{\text{ if }}~x\in X~,\\0~&{\text{ if }}~x\notin X~.\end{cases}}} $$
Question
Is this correct? Note that I am not used to writing a rigorous proof in English so some notations might be awkward.