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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?

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$$ \delta_{f(x)}(\sigma) = \int_{y\in\sigma} d\delta_{f(x)}(y) = \int_{z \in \mathbb{R}} \int_{y \in \sigma} d\delta_{f(z)}(y) d\delta_{x}(z) \tag1 $$

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 $f^{-1}(\sigma)$.

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  • $\begingroup$ Does $f$ need to be continuous for this to hold? I am happy to ask this as a separate question if you prefer. $\endgroup$
    – gigalord
    Jan 23 '20 at 0:35

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