# If $R$ is a $\sigma -$ring then {$E\in X|E\in R$ or $E^c \in R$ } is a $\sigma$ -algebra

If R is $$\sigma -$$ring then {$$E\in X|E\in R \text { or} \ E^c \in R$$ } is a $$\sigma$$ -algebra

$$\sigma -$$ring : A family of sets $$R\in P(X)$$ is called a $$\sigma$$-ring if it is closed under countable unions and differences of 2 sets.

$$\sigma-$$ algebra: A family of sets $$R\in P(X)$$ is called a $$\sigma-$$algebra if it is closed under countable unions and every element has a complement in it.

I had proved that it is enough to show that if $$X\in R$$ then $$R$$ is a $$\sigma-$$ algebra

Case 1) If $$E,E^c\in R$$ then by union we are done.

Case 2) If only either $$E$$ or $$E^c \in R,$$

then I am not able to proceed.

I wanted to solve this. Please give me only a hint.

• Can you rephrase the question, what is the meaning of "if...then...then..."? Feb 6, 2019 at 12:04
• @CalvinKhor Sir SOrry I had done Feb 6, 2019 at 12:08

Let $$\mathcal A:=\{E\in\wp(X)\mid E\in\mathcal R\text{ or }E^{\complement}\in\mathcal R\}$$ where $$\mathcal R$$ denotes a $$\sigma$$-ring on set $$X$$.

Then it is immediate that $$\mathcal A$$ is closed under complements.

Now observe that a $$\sigma$$-ring is closed under countable intersections.

This because we can write: $$\bigcap_{n=1}^{\infty}R_n=R_1\setminus\bigcup_{n=2}^{\infty}(R_1\setminus R_n)$$

Now let $$A_n\in\mathcal A$$ for $$n=1,2,\dots$$ and let it be that $$A_n\in\mathcal R$$ for $$n\in I$$ and $$A_n^{\complement}\in\mathcal R$$ for $$n\in J$$ where we take $$I$$ and $$J$$ to be disjoint and covering subsets of $$\mathbb N$$.

Then $$P:=\bigcap_{n\in I}A_n\in\mathcal R$$ and $$Q:=\bigcup_{n\in J}A_n^{\complement}\in\mathcal R$$.

Consequently $$\bigcap_{n=1}^{\infty}A_n=P\setminus Q\in\mathcal R\subseteq\mathcal A$$.

Proved is now that $$\mathcal A$$ is closed under countable intersections, and combined with the fact that it also closed under complements it can be proved now that $$\mathcal A$$ is closed under countable unions: $$\bigcup_{n=1}^{\infty}A_n=\left(\bigcap_{n=1}^{\infty}A_n^{\complement}\right)^{\complement}$$.

Proved is now that $$\mathcal A$$ is a $$\sigma$$-algebra.

• Sorry Sir , But I am not understanding WHy A is closed under complement ? From Defination it is not direct I think. Feb 6, 2019 at 13:10
• Under what condition do we have $A\in\mathcal A$? Under condition $A\in\mathcal R\vee A^{\complement}\in\mathcal R$. Under what condition do we have $A^{\complement}\in\mathcal A$? Under condition $A^{\complement}\in\mathcal R\vee (A^{\complement})^{\complement}\in\mathcal R$. Now observe that both conditions are exactly the same because $(A^{\complement})^{\complement}=A$. If the condition is satisfied then both $A$ and $A^{\complement}$ are elements of $\mathcal A$. If not then both are not elements of $\mathcal A$. So $A\in\mathcal A\iff A^{\complement}\in\mathcal A$. Feb 6, 2019 at 13:16
• Ohh SO sorry Sir I misunderstood that. Please See my argument : $A\cup A^c=X\in R$ Now $X\A\in R$ this implies closed under complement and cpomtable union is direct . SO we have sigma algebra. THANKS A LOT Feb 6, 2019 at 13:21
• Careful: is not proved that $\mathcal R$ is closed under complementation, but that $\mathcal A$ is closed under complementation. Further it is not direct that $\mathcal A$ is closed under countable unions (we need a proof for that, and I gave one). We have $X=A\cup A^{\complement}\in\mathcal A$ but not necessarily $X=A\cup A^{\complement}\in\mathcal R$ as you seem to think. Feb 6, 2019 at 13:25
• Thanks a lot again Sir. Actually I had missed all crux of question . Now cleared...... Feb 6, 2019 at 13:33