Measurable Spaces – Problem (46/365)

The rest of the chapter on $\sigma$ -algebras goes through the construction of various other measurable spaces such as those on the space of 1) continuous functions, 2) functions continuous on the right, and 3) direct products of measurable spaces. The next chapter introduces methods of introducing probability measures on measurable spaces.

The way we do this is to start with a distribution function $F(x)$ from which we derive a unique probability measure where $P(a,b]) = F(b) - F(a)$ . Here is a problem.

Let $F(x) = P(-\infty, x)$ , then verify that $P(a,b] = F(b) - F(a)$ .

$\displaystyle P(a,b] = P\left( (-\infty,b] \setminus (-\infty, a] \right) \\ = P(-\infty, b] - P(-\infty, a] \text{ by additivity} \\ = F(b) - F(a)$

Verify that $P(a,b) = F(b-) - F(a)$ where $F(x-) = \lim_{y \uparrow x} F(y)$ .

$\displaystyle P(a,b) = P \left( \cup_{n=1}^\infty P(a,b-\frac{1}{n}] \right) \\ = \lim_n P(a,b - \frac{1}{n}] \text{ because } P \text{ is countably additive over } \mathcal{B}(R) \\ = \lim_n F(b - \frac{1}{n}) - F(a) \\ = F(b-) - F(a)$

The proof for the following are similar: $P[a,b] = F(b) - F(a-)$ , $P[a,b) = F(b-) - F(a-)$ , and $P\{x\} = F(x) - F(x-)$ .

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