I am studying for qualifying exams. The question I am struggling with reads as follows:
Consider the probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Take a measurable function $Z \colon \Omega \to [0, \infty)$ with $\text{E}(Z) = 1$. Define a function $\mathbb{Q} \colon \mathcal{F} \to \mathbb{R}$ as $\mathbb{Q}(A) = \text{E}(Z 1_A)$ for all $A \in \mathcal{F}$. Prove that $\mathbb{Q}$ is also a probability measure, using the Lebesgue dominated convergence theorem.
The dominated convergence theorem reads as follows:
Suppose $X_n \to X$. If there exists a random variable $Y \geq 0$ with $\text{E}(Y) < \infty$ such that $\vert X_n \vert \leq Y$ almost surely for all $n$, then $\text{E}(X_n) \to \text{E}(X)$.
To prove that $\mathbb{Q}$ is a valid probability measure, I know I need to prove that
- $\mathbb{Q}(\Omega) = 1$
- $\mathbb{Q}(\bigcup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty} \mathbb{Q}(A_n)$ for disjoint $A_n$
The first task seems simple enough: $\mathbb{Q}(\Omega) = \text{E}(Z 1_{\Omega}) = \text{E}(Z) = 1$. However, the second part is throwing me for a loop. What I have so far is this:
$$ \mathbb{Q}\left(\bigcup_{n=1}^{\infty} A_n \right) = \text{E}\left(Z 1_{\bigcup_{n=1}^{\infty} A_n}\right) = \text{E} \left(Z \left(\sum_{n=1}^{\infty} 1_{A_n}\right) \right) = \text{E}\left( \sum_{n=1}^{\infty} Z 1_{A_n} \right) = \sum_{n=1}^{\infty} \text{E}(Z 1_{A_n}) = \sum_{n=1}^{\infty} \mathbb{Q}(A_n) $$
I have the desired result (or at least I think I have obtained it) without using the Lebesgue Dominated Convergence Theorem. Is there an issue with my proof, or perhaps I need to use the DCT to ensure what I have done is possible? Any guidance is greatly appreciated!
Best Answer
You have used the DCT implicitly in the step: \begin{align} E\left[\sum_{n = 1}^\infty ZI_{A_n}\right] = \sum_{n = 1}^\infty E[ZI_{A_n}]. \end{align}
To see why, note that the infinite series $\sum_{n = 1}^\infty ZI_{A_n}$ is by definition the limit of the partial sum sequence $X_N := \sum_{n = 1}^NZI_{A_n}$ as $N \to \infty$. Because $A_1, A_2, \ldots$ are disjoint, $|X_N| \leq |Z|$ for each $N$, i.e., $\{X_N\}$ is dominated by $Z$, hence the DCT gives: \begin{align} & E\left[\sum_{n = 1}^\infty ZI_{A_n}\right] \\ =& E\left[\lim_{N \to \infty}X_N\right] \\ =& \lim_{N \to \infty}E[X_N] \tag{Apply DCT here} \\ =& \lim_{N \to \infty}\sum_{n = 1}^N E[ZI_{A_n}] \\ =& \sum_{n = 1}^\infty E[ZI_{A_n}]. \end{align}