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I am taking a probability theory course, and I need help understanding a homework question.

Let $(\Omega , \mathcal{F}, \mu)$ be a measure space, and $f:\Omega\rightarrow[0,\infty)$ be such that $\int{f\:\text{d}\mu} = 1$. Show that $v(E) := \int\mathbb{I}_E(\omega)f(\omega)\mu(\text{d}\omega), \: E \in \mathcal{F}$ defines a probability distribution on $(\Omega,\mathcal{F})$.

I went to speak to my professor, and he pointed me towards the definitions of a probability measure and probability space in my textbook. We also covered Lebesgue integration, but I do not know if this applies here. Also, there was some talk about converting these integrals to sums, but I'm quite lost. If anyone can point me in the right direction with explanations, or provide readings or references, I would be grateful.

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up vote 1 down vote accepted

You should check that $v$ satisfies the definition of a probability measure/distribution. That is, you should check that

1) $v$ is a mapping from $\mathcal{F}$ to $[0,1]$.

2) $v(\emptyset)=0$

3) For any sequence of disjoint sets $(A_n)_{n\in\mathbb{N}} \subseteq \mathcal{F}$ the following should hold: $$ v\left(\bigcup_{n\in\mathbb{N}} A_n\right)=\sum_{n\in\mathbb{N}} v(A_n) $$

4) $v(\Omega)=1$

Let me know if any of these causes problems.

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Thank you, this will give me a place to start. It is similar to how a probability space was defined in my text. I will get back to you! –  ialm Sep 29 '12 at 15:10
I think I understand now. 2) follows because the indicator function over the null set is just 0. 4) follows by definition of the function f in the question. 3) follows because the indicator function over union of disjoint sets is just the sum of the indicator functions over the disjoint sets, summation term can come out of the integral. 1) follows because the function is monotone increasing, and 2) and 4) define the smallest and highest values the function can take. –  ialm Sep 30 '12 at 3:32
@iakl: Exactly! –  Stefan Hansen Sep 30 '12 at 9:11
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Probability theory is basically a part of measure and integral theory. Integrals in measure theory are mostly defined using some approximation by step functions (i.e. which take countable many values only, each one over a measurable set, so these integrals are easily computed), and in this generality is called Lebesgue integral.

If $\Omega$ (or the given $\sigma$-algebra of the measurable sets) is finite/countable, then this integral always goes to a concrete sum.

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