Measure theory problems. Prove or disprove the following:
a)If $\mathscr A$ is a $σ$-Algebra on $Ω$ then {$Ω$ \ $A$ : $A$ element of $\mathscr A$} too.
b)A $σ$-Algebra with 3 elements exists.
c)A measure $μ$ on $P(\mathbb R)$ with $μ$({x}}=$1$ exists.

a) Isn't this equal to Ω \ $\mathscr A$? If so then this is the second property of a $σ$-Algebra.
b) Ω={1,2,3}, $\mathscr A$={{1},{2},{3}} is right? 
c) I don't know I just know that for a Lebesgue-Measure $μ$({x}} is equal to $0$. Don't know if any measure exists that can satify c).
 A: a) A $\sigma$-algebra is stable under complementation, hence $\{\Omega\setminus A,A\in\mathcal A\}=\mathcal A$ (it cannot be equal to $\Omega\setminus \mathcal A$, because $\mathcal A$ is a class of subsets of $\Omega$, while $\Omega$ is a set. 
b) Your $\mathcal A$ is not a $\sigma$-algebra (it should at least contain $\Omega$). If $\mathcal A$ is a $\sigma$-algebra with more than two elements, there is $A\in\mathcal A$ which is neither the whole set nor empty. But this is also the case for $\Omega \setminus A$.
c) Consider the counting measure (which assigns to each sets its number of elements if the set if finite, and $+\infty$ otherwise). 
A: a) You need to show that ${\cal B} = \{ \Omega \setminus A | A \in {\cal A} \}$ is a $\sigma$-algebra. This is a straightforward verification using the definitions.
b) A $\sigma$-algebra must be closed under complementation, so, for example,  the set $\{2,3\}$ must be in ${\cal A}$ above. So the ${\cal A}$ above cannot be a $\sigma$-algebra.
In general a $\sigma$-algebra ${\cal A}$ must contain $\emptyset, \Omega$, so $|{\cal A}| \ge 2$. If ${\cal A}$ contains another set $A$, then it must also contain the complement. What does that say about $|{\cal A}|$?
c) The Lebesgue measure cannot work, as any point has measure $0$. Try  letting $\mu A = \begin{cases} 1, &  x \in A \\
0, & \text{otherwise} \end{cases}$ and check that it satisfies the definition.
