Are equations correct? counterexamples Check if these equations are correct: 
(function is measurable) 


)  $ f((A)\bigcup(B) ) = f(A) \bigcup f(B)$
) $ f((A)\bigcap(B)) = f(A) \bigcap f(B)$


I took counterexamples. For the second case i took  $f(x)= x^2$ and $ A=[-1;2], B=[-2;3]$ then $(A\bigcap B)= [-1;2] , f(A\bigcap B) = [1;4] , f(A)=[1;4], f(B)=[4;9] $ so $f(A\bigcap B) \neq f(A) \bigcap  f(B) = [4]$
For the first case i took the same function and i get $A\bigcup B = [-2;3]$    then $ f(A\bigcup B) = [4;9]$ and $ f(A)=[1;4], f(B)=[4;9]  $  but 
$f(A)\bigcup f(B)= [1;9]$ . So again this equation is not correct.  Am i right?
In general, 

$$
f(A \cup B) = f(A) \cup f(B)
$$

and

$$
f(A \cap B) \subseteq f(A) \cap f(B)
$$

For the first one: if $y \in f(A \cup B)$, either $y$ is the image of some element in $A$, or the image of some element in $B$, so respectively $y\in f(A)$ or $y\in f(B)$, which means that $f(A \cup B) \subseteq f(A) \cup f(B)$. Reciprocally , if $y \in f(A) \cup f(B)$, either $y\in f(A)$ or $y\in f(B)$, and let's assume without loss of generality that $y\in f(A)$. Since $A \subseteq A \cup B$, $y$ is the image of an element of $A \cup B$, and therefore we obtain $f(A) \cup f(B) \subseteq f(A \cup B)$. This completes the proof.

As for the second one, if $y \in f(A\cap B)$, $y = f(x)$ with $x\in A\cap B$. We can immediately tell that $x$ is both in $A$ and in $B$. So $y$ is in $f(A)$ and $f(B)$ respectively and therefore in $f(A) \cap f(B)$, which  proves the claim.

Your second example proves that the equality in the second case does not always hold, and I'll leave you to examine why the first one is not an actual counterexample.
You need to know that your function is isomorphism injection,byjection,continuous)otherwise you can't speak about equalities