# Measure Theory – Proof that the intersection of Sigma algebras is a Sigma algebra

For $$A subset 2 ^ { Omega}$$,

let $$sigma (A): = bigcap _ { mathcal {F} in {F}} mathcal {F}$$,

or $$F = { mathcal {F} subset 2 ^ { Omega}: {F}$$ is a $$sigma$$-Algebra and $$A subset mathcal {F} }.$$

The goal is to prove that $$sigma ({A})$$ is a $$sigma$$-algebra.

It seems to me that we can define $$sigma {A}$$ as $$bigcap _ { mathcal {F} _ {k} in F} mathcal {F} _ {k} forall k in K$$.

Then if $$A in mathcal {F}$$, $$A in mathcal {F} _ {k}$$ for everyone $$k$$And the same is true $$A_ {c}$$.

So, to show that $$sigma ({A})$$ is a sigma algebra, we simply show the following: that if $$A_ {n} in sigma ({A})$$ for everyone $$n = 1,2, dots$$,

then $$A_ {n} in mathcal {F} _ {k}$$ for everyone $$n, k$$, and

$$bigcup_ {n = 1} A_ {n} in mathcal {F} _ {k} forall k$$,

which implies $$bigcup_ {n = 1} A_ {n} in mathcal {F}$$.

The only thing that puts me off, is that we have $$A subset mathcal {F}$$, do not $$A in mathcal {F}$$.

Is it true that if A, F are both $$subset$$ $$2 ^ { Omega}$$, and $$A subset F$$, this $$A in F$$? I think that's true, but does it have to be proven?