# Thread: Show intersection of sigma-algebras is again a sigma-algebra

1. ## Show intersection of sigma-algebras is again a sigma-algebra

Hi

Let $(\mathcal{G}_{\alpha})_{\alpha \in A}$ be an arbitrary familiy of $\sigma$-algebras defined on an abstract space $\Omega$. Show that $\mathcal{H} = \cap_{\alpha \in A} \mathcal{G}_{\alpha}$ as also a $\sigma$-algebra.

Solution(?)

Since $\left\{\emptyset\right\},\left\{\Omega\right\} \in \mathcal{G}_{\alpha},\alpha \in A$, we have that $\cap_{\alpha \in A} \mathcal{G}_{\alpha} = \mathcal{H}$ as always "at least" $\left\{\emptyset,\Omega\right\}$, which is the trivial $\sigma$-field.

In fact, if $\mathcal{G}_{\alpha_{i}} \cap \mathcal{G}_{\alpha_{j}} = \emptyset$, $\alpha_{i},\alpha_{j} \in A$, then $\mathcal{H} = \left\{\emptyset,\Omega\right\}$.

Otherwise, if say $A \subset \Omega$ is in $\mathcal{G}_{\alpha}$, $\alpha \in A$, then $A^{c} \in \mathcal{G}_{\alpha}$ $\Rightarrow A,A{c} \in \mathcal{H}$, and $\mathcal{H}$ is a $\sigma$-algebra.

2. Looks like you have the idea, but your notation is a little off. For example you mean $\emptyset \in \mathcal G_{\alpha}$.

In your last line you can't use A - it's already being used for the index set. Perhaps use X instead. Your last line should look something like this:

$X\in \mathcal H \Leftrightarrow (\forall \alpha \in A) X\in G_{\alpha} \Leftrightarrow (\forall \alpha \in A) \Omega - X \in G_{\alpha} \Leftrightarrow \Omega - X \in \mathcal H$.

Of course you still have to check closure under countable unions and intersections, but each part of the proof is quite straightforward.

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# union of two subgrl is again nd again a subgr in hindi

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