show/hide this revision's text 2 Replace $\bar{\mathcal{B}}$ by $\mathcal{L}$. Some other amplifications.

Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$, so the Lebesgue $\sigma$-algebra $\bar{\mathcal{B}}$ \mathcal{L}$ is its completion . with respect to Lebesgue measure $m$. Then for functions $f : \mathbb{R} \to \mathbb{R}$, "Borel measurable" means $(\mathcal{B}, \mathcal{B})$-measurable. "Lebesgue measurable" means $(\bar{\mathcal{B}},\mathcal{B})$ (\mathcal{L},\mathcal{B})$ measurable; note the asymmetry! Already this notion has some defects; for instance, if $f,g$ are Lebesgue measurable, $f \circ g$ need not be, even if $g$ is continuous. (See Exercise 2.9 in Folland's Real Analysis.)

$(\bar{\mathcal{B}}, (\mathcal{L}, \bar{\mathcal{B}})$-measurable mathcal{L})$-measurable functions are not so useful; for instance, a continuous function need not be $(\bar{\mathcal{B}}, (\mathcal{L}, \bar{\mathcal{B}})$-measurable. mathcal{L})$-measurable. (The $g$ from the aforementioned exercise is an example.) $(\mathcal{B}, \mathcal{L})$ is even worse.

Given a probability space $(\Omega, \mathcal{F},P)$, our random variables are $(\mathcal{F}, \mathcal{B})$-measurable functions $X : \Omega \to \mathbb{R}$. The Lebesgue $\sigma$-algebra $\mathcal{L}$ does not appear. As mentioned, it would not be useful to consider $(\mathcal{F}, \bar{\mathcal{B}})$-measurable mathcal{L})$-measurable functions; there simply may not be enough good ones, and they may not be preserved by composition with continuous functions. Anyway, the right analogue of "Lebesgue measurable" would be to use the completion of $\mathcal{F}$, \mathcal{F}$ with respect to $P$, and this is commonly done. Indeed, many theorems assume a priori that $\mathcal{F}$ is complete.

Note that, for similar reasons as above, we should expect $f(X)$ to be another random variable when $f$ is Borel measurable, but not when $f$ is Lebesgue measurable. Using $(\mathcal{F}, \bar{\mathcal{B}})$ mathcal{L})$ in our definition of "random variable" would not avoid this, either.

I think the

The moral is that this: To get as many $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable functions $f : X \to Y$ as possible, one wants $\mathcal{B}_X$ to be as large as possible, so it makes sense to use a complete $\sigma$-algebra is nice to have on there. (You already know some of the domainnice properties of a this, e.g. an a.e. limit of measurable functionfunctions is measurable.) But one wants $\mathcal{B}_Y$ to be as small as possible. It When $Y$ is not necessarily so nice a topological space, we usually want to have on be able to compose $f$ with continuous functions $g : Y \to Y$, so $\mathcal{B}_Y$ had better contain the codomainopen sets (and hence the Borel $\sigma$-algebra), but we should stop there.
show/hide this revision's text 1

One should be careful with the definitions here. Notation: Given measurable spaces $(X, \mathcal{B}_X), (Y, \mathcal{B}_Y)$, a measurable map $f : X \to Y$ is one such that $f^{-1}(A) \in \mathcal{B}_X$ for $A \in \mathcal{B}_Y$. To be explicit, I'll say $f$ is $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable.

Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$, so the Lebesgue $\sigma$-algebra $\bar{\mathcal{B}}$ is its completion. Then for functions $f : \mathbb{R} \to \mathbb{R}$, "Borel measurable" means $(\mathcal{B}, \mathcal{B})$-measurable. "Lebesgue measurable" means $(\bar{\mathcal{B}},\mathcal{B})$ measurable; note the asymmetry! Already this notion has some defects; for instance, if $f,g$ are Lebesgue measurable, $f \circ g$ need not be, even if $g$ is continuous. (See Exercise 2.9 in Folland's Real Analysis.)

$(\bar{\mathcal{B}}, \bar{\mathcal{B}})$-measurable functions are not so useful; for instance, a continuous function need not be $(\bar{\mathcal{B}}, \bar{\mathcal{B}})$-measurable. (The $g$ from the aforementioned exercise is an example.)

Given a probability space $(\Omega, \mathcal{F},P)$, our random variables are $(\mathcal{F}, \mathcal{B})$-measurable functions $X : \Omega \to \mathbb{R}$. The Lebesgue $\sigma$-algebra does not appear. As mentioned, it would not be useful to consider $(\mathcal{F}, \bar{\mathcal{B}})$-measurable functions; there simply may not be enough good ones. Anyway, the right analogue of "Lebesgue measurable" would be to use the completion of $\mathcal{F}$, and this is commonly done. Indeed, many theorems assume a priori that $\mathcal{F}$ is complete.

Note that, for similar reasons as above, we should expect $f(X)$ to be another random variable when $f$ is Borel measurable, but not when $f$ is Lebesgue measurable. Using $(\mathcal{F}, \bar{\mathcal{B}})$ in our definition of "random variable" would not avoid this, either.

I think the moral is that a complete $\sigma$-algebra is nice to have on the domain of a measurable function. It is not necessarily so nice to have on the codomain.