A Haar measure is a left invariant Borel measure which is not identically zero A Haar measure is a Borel measure $\mu$ in a locally compact topological group $X$, such that $\mu$(U)>0 for every non empty Borel open set $U$, and $\mu(xE)$=$\mu(E)$ for every Borel set $E$.
A Borel measure is a measure $\mu$ defined on the class $S$ of all Borel sets and such that $\mu$(C)< $\infty$ for every C in $C$.
we shall denote by $C$ the class of all compact subsets of X ,by S the $\sigma$-ring generated by $C$.we shall call the sets of S the borel sets of X.
We want to show that the first property is equivalent to the assertion that $\mu$ is not identically zero.
If $\mu(U)$=0 for some non empty Borel open set $U$, and if $C$ is any compact set,then the class $\{xU\mid x \in C\}$ is an open covering of $C$.
Since $C$ is compact, there exists a finite subset $\{x_1,\dots,x_n\}$ of $C$ such that $C \subset \bigcup\limits_{i=1}^{n}x_i U$, and the left invariance of $\mu$ implies that 
$$\mu(C)\leq\sum_{i=1}^{n}\mu(x_iU)=n\mu(U)=0.
$$ Since the vanishing on the class $C$ of all compact sets implies its vanishing on the class $S$ of all Borel sets, we obtain the desired result.  
My question is:

Why is the class $\{xU\mid x \in C\}$ an open covering of $C$?
  And why does the vanishing on the class $C$ of all compact sets implies its vanishing on the class $S$ of all Borel sets?

Thanks very much in advance.
 A: Let me begin by clarifying a few things about the definition.  First, I assume from your phrasing that you are defining the "Borel sets" to be the $\sigma$-ring generated by the compact subsets of $X$ (rather than the $\sigma$-algebra generated by the open subsets of $X$).  Second, the definition of Haar measure should not say that $\mu(C)<\infty$ for all Borel sets $C$; it should only say that $\mu(C)<\infty$ for all compact sets $C$.  If you assume $\mu(C)<\infty$ for all Borel sets, this actually implies $X$ must be compact (see this question).
Let me now address your questions.  For the first question, it is true that $\{xU\mid x\in C\}$ covers $C$ if $U$ contains the identity $1\in G$, since then $x\in xU$ for each $x\in C$.  But the $U$ we were given might not contain the identity, and in that case there is no reason to expect $\{xU\mid x\in C\}$ to cover $C$.  To fix this, just let $y\in U$ be any point, and replace $U$ with $y^{-1}U$.
For the second question, simply note that the collection of Borel sets $E$ such that $\mu(E)=0$ is a $\sigma$-ring, so if it contains all compact sets, then by definition it contains all Borel sets.
