I'm read a measure theory book and it says:

Every outer measure induces $\mu$ its own $\sigma$-algebra of $\mu$-measurable sets and $\mu$ is countably additive on this $\sigma$-algebra. However, the $\sigma$-algebra of measurable sets strongly depends on $\mu$. Consider the following examples:

If $\mu = 1$ when $0 \in A$ and $0$ otherwise (i.e. dirac delta measure), then every subset of $\mathbb{R}$ is $\mu$-measurable. On the other hand, if $\mu(\emptyset) = 0$ and $\mu(A) = 1$ when $A \neq \emptyset$ the $\sigma$-algebra of measurable sets reduces to $\{0, \mathbb{R}\}$.

I'm having trouble verifying that these measures generate these sigma algebras. Thoughts?


Recall that $E$ is $\mu$-measurable iff

$$\forall A \subseteq X: \mu(A) = \mu(A \cap E) + \mu (A \cap E^\complement)\tag{1}$$

Dirac measure $\mu_0$: suppose $E$ is any subset of $\mathbb{R}$.

Let $A \subseteq \mathbb{R}$ be arbitrary.

If $0 \in A$ then $(1)$ reduces to $1= \mu(A \cap E) + \mu(A \cap E^\complement)$ and as $0$ is in exactly one of $E$ or $E^\complement$ the right hand side has one $0$ and $1$, so their sum $1$ too.

If $0 \notin A$, all $3$ sets in $(1)$ have measure $0$, and $(1)$ checks out too.

As $A$ was arbitrary, $E$ is $\mu$-measurable.

$\mu$ the trivial $0$-$1$ measure:

taking $E=\mathbb{R}$ or $E=\emptyset$ reduces $(1)$ to $0=0+0$ for $A=\emptyset$ or $1=1+0$ for other $A$. So always $\emptyset, \mathbb{R}$ are $\mu$-measurable.

If $\emptyset \neq E \neq \mathbb{R}$, let $p \in E, q \notin E$ and define $A=\{p,q\}$, then $(1)$ for this $A$ reduces to $1=1+1$ (all sets are non-empty so have measure $1$) so this fails for this $A$. Ergo, $E$ is not measurable, and we have that the measurable sets are only $\{\emptyset,\mathbb{R}\}$.


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