Let $X$ a set and $\delta_0:\mathcal P(X)\longrightarrow \mathbb R$ defined by $$\delta_0(A)=\begin{cases}1&0\in A\\ 0&sinon\end{cases}\quad A\subset X.$$
We recall that $\delta_0$ is a measure called "Dirac measure".
Let $f:X\longrightarrow \mathbb R$ a function. What does mean "$f$ is a $\delta_0-$ measurable function" ?
Attempt : Normaly, for function $$f:(X,\mathcal M,\mu)\longrightarrow (Y,\mathcal N,\nu),$$ $f$ is measurable if $$f^{-1}(N)\in \mathcal M$$ for all $N\in\mathcal N$. The problem is that in my question no $\sigma -$algebra are definite in any set (and I suppose that I have to find such $\sigma -$algebra).
Conclusion : I suspect that the tribute on $X$ is $\mathcal P(X)$ (the poxer set of $X$), and it's measurable if $$f^{-1}(B)\in \mathcal P(X),$$ for all $B$ borealian. But with such definition, all function are $\delta_0$ measurable... so may be it's not correct ?