Is a measure measurable?

Question

This could totally be a stupid question but I'm unsure: is a measure (ie positive, countable additive on a $\sigma$ algebra, 0 for the empty set) actually a measurable function (wrt to the Borel-sigma algebra on $\mathbb{R}$)?

Answer 1

Given a function, a $\sigma$-algebra on the domain and a $\sigma$-algebra on the codomain, you can ask whether the function is measurable with respect to the given $\sigma$-algebras. You have only specified a $\sigma$-algebra on the codomain, so we can not answer your question.

What probably confused you is the fact that the domain of a measure is a $\sigma$-algebra on some set, so we need to consider a $\sigma$-algebra on another $\sigma$-algebra - the domain of the measure - to determine whether it is measurable.

Answer 2

For a measure space $(X,\mathcal{F},\mu)$ you have that

$$ \mu^{-1}((-\infty ,c])=\{A\in \mathcal{F}:\mu(A)\leqslant c\} $$

so you will need a $\sigma $-algebra defined in $\mathcal{F}$ to define the measurability of $\mu$. Well, you can define this $\sigma $-algebra using $\mu$, this will give an induced $\sigma $-algebra in $\mathcal{F}$, and we can note it by $\sigma (\mu)$.

Answer 3

Given two sets $X_1,X_2$ with $\sigma$ algebras $\Sigma_1,\Sigma_2$ the function $f:X_2 \to X_2$ is called measurable if the preimage of each measurable set in $X_2$ is measurable in $X_1$.

However a measure on $X_1$ is not a function on $X_1$. It does not take points of $X_1$ as inputs. It takes subsets as inputs. So it does not make sense to ask whether it is a measurable function or not.

Answer 4

If $X$ is a seperable Hausdorff space, denote by $\mathscr{Bor}(X)$ its Borel $\sigma$-algebra. If $\mu$ is a $\sigma$-finite positive measure on $\mathscr{Bor}(X)$. Then $x \in X \mapsto \mu (\{x\}) \in \overline{\mathbb{R}}_{+}$ is $(\mathscr{Bor}(X), \mathscr{Bor}(\overline{\mathbb{R}}_{+}))$-measurable.