Measure-theoretic probability, push-forward measure

Question

I cannot make sense of the definition of the push-forward measure. According to my script, the domain of $\mathbb{P}_{X}$ is $\left\{ A\subset\mathbb{R}:X^{-1}\left(A\right)\in\mathcal{F}\right\} $ where $\mathcal{F}$ is the $\sigma$-algebra.

Here is what I don't get:

Since $\mathcal{F}$ is a $\sigma$-algebra, it must contain the outcome space. Since it must be closed under complementation, it must contain the empty set.

Let's say I have a set $A$ with values that are not in the range of $X$. Then, $X^{-1}(A)$ is the empty set. Since the empty set is in $\mathcal{F}$, that $A$ is in the domain of $\mathbb{P}_{X}$. So basically anything is in the domain. Where am I wrong?

Answer 1

The set function $\mathbb{P}_{X}$ is indeed a probability measure.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $X:\Omega\to\mathbb{R}$ be a random variable.

In order to solve the problem, remember the definition of a probability measure:

• One must have that $\mu(\Omega) = 1$,
• One must also have that $\mu(A)\geq 0$, for every $A\in\mathcal{F}$,
• And the set function $\mu$ must be $\sigma$-additive.

Notice that all these restrictions are satisfied by $\mathbb{P}_{X}$ defined over $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

Here $\mathcal{B}(\mathbb{R})$ denotes the Borel $\sigma$-algebra: the $\sigma$-algebra generated by the open sets of $\mathbb{R}$.

Indeed, the first statement is true, because \begin{align*} \mathbb{P}_{X}(\mathbb{R}) = \mathbb{P}(X^{-1}(\mathbb{R})) = \mathbb{P}(\Omega) = 1. \end{align*}

Similarly, the second statement is true as well. Let $A\subseteq\mathbb{R}$. Then we have that \begin{align*} \mathbb{P}_{X}(A) = \mathbb{P}(X^{-1}(A))\geq 0 \end{align*} since $X^{-1}(A)\in\mathcal{F}$.

Finally, let $A_{n}$ be a countable collection of subsets from $\mathcal{B}(\mathbb{R})$ which are pairwise disjoint.

Moreover, let $B_{n} := X^{-1}(A_{n})$ (which implies necessarily that $B_{n}$ are pairwise disjoint as well).

Then it results the $\sigma$-additivity of $\mathbb{P}_{X}$: \begin{align*} \mathbb{P}_{X}\left(\bigcup_{n=1}^{\infty}A_{n}\right) & = \mathbb{P}\left(X^{-1}\left(\bigcup_{n=1}^{\infty}A_{n}\right)\right)\\ & = \mathbb{P}\left(\bigcup_{n=1}^{\infty}X^{-1}(A_{n})\right)\\ & = \mathbb{P}\left(\bigcup_{n=1}^{\infty}B_{n}\right)\\ & = \sum_{n=1}^{\infty}\mathbb{P}(B_{n})\\ & = \sum_{n=1}^{\infty}\mathbb{P}(X^{-1}(A_{n}))\\ & = \sum_{n=1}^{\infty}\mathbb{P}_{X}(A_{n}) \end{align*}

and we are done.