Laplace functional of sum of independent uniformly distributed random variables

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I'm doing some of the exercises in Cinlar's "Probability and Stochastics" to better understand the material. This exercise (VI.1.17) is taken from page 247:

Fix an integer $n \geq 1$. Let $X_1,\ldots,X_n$ be independent and uniformly distributed over $E = [0,1]$. Define M by $$Mf(\omega) = \sum_i f \circ X_i(\omega), \qquad \omega \in \Omega, $$ where $f$ is a Borel function on E. Note that this makes M a random measure on $(E,\mathcal{E})$. Compute the Laplace functional of M.

The Laplace functional is defined as a mapping $$f \mapsto \mathbb{E}[e^{-Mf}].$$ I'm having a little trouble calculating it, here's my current approach:

$$\begin{align}\mathbb{E}[e^{-Mf}] &= \mathbb{E}[e^{\sum_{i=1}^n -f(X_i)}]\\ &= \mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]\end{align}$$ I think the next step would be to use independence to put the product outside, and then "insert" the uniform distribution - but in the book it's not really mentioned what the uniform distribution "means", so I don't know what exactly to insert.

Some good hints or the next steps of the calculation would be appreciated.

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I'm assuming:

  1. $f$ is not a Borel set but a Borel-measurable function $\mathbb R \to \mathbb R$.

  2. $\mathcal{E} = \mathscr{M}(E) \ or \ \mathscr{B}(E)$


$$\mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]$$

$$= \prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

Now

$$\mathbb{E}[e^{-f(X_i)}] = \int_{\Omega} e^{-f(X_i)} d\mathbb P$$

$$= \int_{E} e^{-f(x_i)} d \mathcal{L}_{X_i}(t)$$

$$= \int_{E} e^{-f(x_i)} f_{X_i}(x_i) dx_i$$

Now insert the uniform distribution $f_{X_i}(x_i) = 1_{E}(x_i)$

$$= \int_{E} e^{-f(x_i)} 1 dx_i$$

$$= \int_{E} e^{-f(x_i)} dx_i$$

Hence,

$$\prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

$$= \prod_{i=1}^n [\int_{E} e^{-f(x_i)} dx_i]$$