# Can we apply the Dirac Measure to itself in the Lebesgue Integral?

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?

$$\delta_{f(x)}(\sigma) = \int_{y\in\sigma} d\delta_{f(x)}(y) = \int_{z \in \mathbb{R}} \int_{y \in \sigma} d\delta_{f(z)}(y) d\delta_{x}(z) \tag1$$
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 $$f^{-1}(\sigma)$$.
• Does $f$ need to be continuous for this to hold? I am happy to ask this as a separate question if you prefer. Jan 23 '20 at 0:35