Measure
A Measure on $(X,\sum)$ is a function $\mu:\sum\rightarrow[0,\infty]$ such that the
following conditions are satisfied:

(i) $\mu(\phi)=0$
(ii) if $A_{i}$ is a sequence of mutually disjoint sets in $\sum$
i.e $A_{i}\epsilon\sum$ for each $i=1,2,3,...$ and ${A_i}\cap{A_j}=\phi$ then for
$i\neq{j}$, $\mu(\cup^\infty_{i=1}A_i)=\sum^\infty_{i=1}(A_i)$ this is the countable
additivity property,
we must know that the measure is also finitely additive.
Measurable space
A measurable space is a pair $(X,\sum)$ where X is a set and $\sum$ is a
$\sigma$-algebra in X. whereby the elements of this algebra are called the measurable sets.
Examples
(i) the counting measure
(ii) The probability measure
(iii) the unit point mass measure.
Counter Examples
(1)The problem of measuring all bounded subsets of a line was done by G.VITALI where he
asserts that for $A\subseteq\mathbb{R}$, we determine for every A a positive number
$\mu(A)$ which must satisfy the following:
(i) for all $x$ in $\mathbb{R}$ all $A\subseteq\mathbb{R}$ then $\mu(A)=\mu(A-x)$.
(ii) for all $\{A_n\subseteq\mathbb{R}:A_n is bounded \}$ a disjoint sequence,
then $\sum^\infty_{n=1}(A_n)=\mu(\cup^\infty_{n=1}(A_n)$.
(iii)the $\mu(0,1)=1$
(2)if $V\subseteq[0,1]$ is a vitali set then v is not lebesgue measurable.