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We are looking at the measure space $(\mathbb{R}^d, \mathcal{P}(\mathbb{R}^d),\delta_p)$, for all $A\subset \mathbb{R}^d$:

$\delta_p(A):=\begin{cases} 1, & \text{if } p \in A \\ 0, & \text{otherwise} \end{cases}$.

  1. Let $A \subset \mathbb{R}^d$ and $f:\mathbb{R}^d \rightarrow \overline {\mathbb {R}}$ arbitrary. Show that $f$ is measurable and calculate $\int_A f \space d \delta_p$.

  2. Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^d$. Does an integratable function $h: \mathbb{R}^d \rightarrow \mathbb{R}$ exist such that for $A \subset \mathbb{R}^d: \delta_p(A)=\int_Ah \space d \lambda$?


$\int_{A}fd\delta_p= \begin{cases} f(p) & \text{if}~ p\in A\\ 0 & \text{otherwise} \end{cases}$ is the integral but how do I show this rigorously?

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  • $\begingroup$ For 1. note that for any set $A$ we must have $f^{-1} (A) \in \mathcal{P}(\mathbb{R}^d)$. $\endgroup$
    – copper.hat
    Nov 30, 2020 at 17:31

1 Answer 1

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Use the fact that, for an increasing sequence $(f_n)_n$ of simple functions with $\lim_{n \to +\infty}f_n(x) = f(x)$, $$\int f~d\delta = \lim_{n \to \infty} \int f_n~d\delta$$ by definition of the integral. (Note that in the definition of the integral we suppose $f \ge 0$, you just have to separate $f^+$ and $f^-$).

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