# Verifying some examples of sigma algebra of measurable sets

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 $$\{\emptyset,\mathbb{R}\}$$.