Say we have some function $f$ on the reals and a Borel set $\sigma$ from the $\sigma$-algebra of $\mathbb{R}$. My understanding from https://en.wikipedia.org/wiki/Dirac_measure is that the Dirac measure
$\delta_x(\sigma) = \begin{cases}
1 & x\in \sigma \\
0 & x\not\in \sigma \\
\end{cases}$
satisfies the following properties
- $f(x) = \int_{y\in\mathbb{R}} f(y) d\delta_{x}(y)$
- $\delta_{f(x)}(\sigma) = \int_{y\in\sigma} d\delta_{f(x)}(y)$
Is it the case that:
\begin{align*}
\delta_{f(x)}(\sigma) = \int_{y\in\sigma} d\delta_{f(x)}(y) = \int_{z \in \mathbb{R}} \int_{y \in \sigma} d\delta_{f(z)}(y) d\delta_{x}(z)
\end{align*}
If not, why?
Best Answer
$$ \delta_{f(x)}(\sigma) = \int_{y\in\sigma} d\delta_{f(x)}(y) = \int_{z \in \mathbb{R}} \int_{y \in \sigma} d\delta_{f(z)}(y) d\delta_{x}(z) \tag1 $$
Yes, all three of these are $$ \begin{cases}1,\quad&\text{if }f(x) \in \sigma\\0,\quad&\text{if }f(x) \notin \sigma\end{cases}, $$ or the indicator function of $f^{-1}(\sigma)$.