I'm little bit confused about Borel measure. What are the check to do for a Borel measure identification? Is true that these functions defined on every Borel set $F \in 2^R$ are Borel?
1) $\mu(F) := \sup F - \inf F$
2) $\mu(F) = 0$ if F countable, $+\infty$ if not
3) $\mu(F) = 0$ if $x \in F$, $\mu(F) = 1$ if not
4) $\mu(F) = $$\int_{F}^{} e^x dx$
Here my ideas for those questions
1) I can define $F=[a,b)$ and since the sup of F is not defined $\rightarrow$ False
2) According to $\sigma$-algebra measure definition, $\mu(\emptyset)=0$, but here the measure is 0 even if it's not empty $\rightarrow$ False
3) It's like the "inverse" of the Dirac measure, or an inverse of the characteristic function, but other than these motivation I can't figure out other stuff $\rightarrow$ True
4) $\mu(F) = $$\int_{F}^{} e^x dx = C\int_{B_r(0)}^{} {\chi}_F \rho^2 e^\rho d\rho$ this integral is finite $\rightarrow$ True
I need more structure on the subject, I'm afraid I'm missing huge chunks of proofs
There are many mistakes in your arguments. '$\sup F$ not defined' is not an argument for 1). If $A$ is the set of all rational numbers in $(0,)$ and $B$ is the set of all irrational numbers in $(0,)$ then $\mu (A)=1, \mu(B)=1$ and $\mu (A \cup B)=1$. Since $A$ and $B$ are disjoint this contraducts the definition of a measure. So the answer to 1) is NO.
There is no requirement that the measure of a non-empty set is not zero.The answer to 2) is YES and I leave it to you to verify that this $\mu$ is indeed countable additive. [Let me know if you need help with this].
In 3) $\mu (\emptyset) =1$ which is not permissible, so the answer is NO.
in 4) $\mu$ is a Borel measure and this follows by standard measure theoretic arguments.