# Is expectation countably additive?

I am supposed to prove that

Let $X \ge 0$ be a random variable defined on $(\Omega, \mathcal{A}, > P)$ and $\mathbb{E}[X] = 1$. Define $Q: \mathcal{A} \to \mathbb{R}$ by $Q(A) = \mathbb{E}[X1_{A_n}]$. Show that Q is a probability measure on $(\Omega, \mathcal{A})$

Of course $Q(\Omega) = 1$. Now let $A_n \in \mathcal{A}$ pairwise disjoint events. Then

$$Q\left(\bigcup A_n\right) = \mathbb{E}\left[X1_{\bigcup A_n}\right] = \mathbb{E}\left[\sum X1_{A_n}\right].$$

Now can I say that $\mathbb{E}[\sum X1_{A_n}] = \sum \mathbb{E}[X1_{A_n}] = \sum Q(A_n)$, completing the proof?

I know that expectation is a linear operator, but how does it behave with countable sum?

• Try to prove the two inequalities instead.. – Snufsan May 17 '14 at 17:06

You can pass to countable sums applying theorems like Monotone Convergence or Dominated Convergence. In your case, the monotone convergence theorems does the job, once the functions defined by

$$f_n = \sum_{i=1}^n X1_{A_i}$$

converge to

$$f = \sum_{i=1}^{\infty} X1_{A_i}$$

monotonically, right?

Then you have

$$\mathbb{E}[f] = \lim_n \mathbb{E}[f_n] = \lim_n \sum_{i=1}^n \mathbb{E}[X1_{A_i}] = \sum_{i=1}^{\infty} \mathbb{E}[X1_{A_i}]$$

• Right. Though one should also take the functions to be positive, right? – Ant Jun 1 '14 at 18:16
• Right! You should have the right hypothesis to apply the monotone convergence theorem. In this case, $X$ Positive. – Rodrigo Ribeiro Jun 1 '14 at 21:40

Expectation is given by the integral, i.e. for a random variable $X$ on a probability space $(Y,M,P)$ we have

$\mathbb{E}X = \int_Y X dP$

Because of $X \geq 0$, you can now apply the monotone convergence theorem to complete your proof.