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$$\mathcal A_1 = \{A \subseteq \mathbb R^d : A\text{ is open in }\mathbb R^d \text { or } \mathbb R^d\setminus A\text{ is open in }\mathbb R^d\}$$

and:

$$\mathcal A_2 = \{A \subset \mathbb R^d : A\text{ is dense in }\mathbb R^d \text{ or } \mathbb R^d\setminus A \text{ is dense in }\mathbb R^d\}$$

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    Please format your question more clearly – 2017-02-26
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    Welcome to math.SE! The way this question is asked makes it seem like an exercise problem. If this is the case, you are much more likely to receive a good answer if you take some time to say how you have thought about this yourself. Asking people to answer your homework questions for you is not received well. – 2017-02-26
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    First, you probably mean to include $\emptyset$ in both of your collections, otherwise the question is pretty trivial. Here's a hint -- when showing something is a $\sigma$-algebra, the meat is in showing the collection is closed under countable unions and compliments. We know that $A_1$ and $A_2$ are closed under countable (in fact, arbitrary) unions. So that really leaves only one thing to check. Try taking compliments of some open sets and dense sets and see what happens. – 2017-02-26
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    @guest E.g. $(0,2]=(0,1)\cup[1,2]$ is a union of sets in $A_1$, but is not an element of $A_1$. Both collections are evidently closed under the formation of complements. Also the empty set is evidently an element of both collections. – 2017-02-26
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    @drhab The question has since been edited. The question, as originally written, had $A_1 = \{A: A$ is open or $A = \mathbb{R}^d\}$ and $A_2 = \{A: A$ is dense or $A = \mathbb{R}^d\}$. My comment was directed at the original question, not the new one. – 2017-02-26

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Take $d=1$ and let $\mathbb R$ be equipped with its usual topology.

Observe that $(-\infty,1]$ and $(0,\infty)$ are both elements of $\mathcal A_1$.

But $(0,1]=(-\infty,1]\cap(0,\infty)\notin\mathcal A_1$, and a $\sigma$-algebra is closed under finite intersections.

Observe that $\mathbb Q\setminus\{q\}\in\mathcal A_2$ for every $q\in \mathbb R$.

But $\mathbb Q\setminus[0,1]=\bigcap_{q\in\mathbb Q\cap[0,1]}\mathbb(Q\setminus\{q\})\notin\mathcal A_2$, and a $\sigma$-algebra is closed under countable intersections.